Python 3, 76 bytes

F=lambda x:x<2or x*F(x-1)

f=lambda x,n=1:x<2or n*(F(x)>F(x-n)**2)or f(x,n+1)

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Relies on the fact that for eating $n$ candies to still win and the total number of candies being $x$, $frac{x!}{(x-n)!}>(x-n)!$ must be true, which means $x!>((x-n)!)^2$.

-1 from Skidsdev

-3 -6 from BMO

-3 from Sparr

+6 to fix x = 1

JavaScript (ES6), 53 bytes

n=>(g=p=>x<n?g(p*++x):q<p&&1+g(p/n,q*=n--))(q=x=1)||n

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Working range

Interestingly, the differences between the kids' products are always big enough that the loss of precision inherent to IEEE 754 encoding is not an issue.

As a result, it works for $0 le n le 170$. Beyond that, both the mantissa and the exponent overflow (yielding +Infinity) and we'd need BigInts (+1 byte).

How?

Let $p$ be the candy product of the other kid and let $q$ be our own candy product.

1. We start with $p=n!$ (all the candy for the other kid) and $q=1$ (nothing for us).



2. 
   

   We repeat the following operations until $qge p$:

   
   
   
   
   
   • divide $p$ by $n$
     
   
   
   • multiply $q$ by $n$
     
   
   
   • decrement $n$
     
   
   
   
   

The result is the number of required iterations. (At each iteration, we 'take the next highest candy from the other kid'.)

Commented

This is implemented as a single recursive function which first compute $n!$ and then enters the loop described above.

n => (           // main function taking n

  g = p =>       // g = recursive function taking p

    x < n ?      //   if x is less than n:

      g(         //     this is the first part of the recursion:

        p * ++x  //     we're computing p = n! by multiplying p

      )          //     by x = 1 .. n

    :            //   else (second part):

      q < p &&   //     while q is less than p:

      1 + g(     //       add 1 to the final result

        p / n,   //       divide p by n

        q *= n-- //       multiply q by n; decrement n

      )          //

)(q = x = 1)     // initial call to g with p = q = x = 1

|| n             // edge cases: return n for n < 2

Jelly, 9 bytes

ḊPÐƤ<!€TL

Try it online! Or see the test-suite.

How?

ḊPÐƤ<!€TL - Link: integer, x                   e.g. 7

Ḋ         - dequeue (implicit range of x)           [   2,   3,   4,   5,   6,   7]

  ÐƤ      - for postfixes [all, allButFirst, ...]:

 P        -   product                               [5040,2520, 840, 210,  42,   7]

      €   - for each (in implicit range of x):

     !    -   factorial                             [   1,   2,   6,  24, 120, 720, 5040]

    <     - (left) less than (right)?               [   0,   0,   0,   0,   1,   1, 5040]

          -   -- note right always 1 longer than left giving trailing x! like the 5040 ^

       T  - truthy indices                          [                       5,   6, 7   ]

        L - length                                  3

APL (Dyalog Unicode), 10 bytes

+/!≤2*⍨!∘⍳

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Port of Dennis' answer. Thanks to, well, Dennis for it.

How:

+/!≤2*⍨!∘⍳ ⍝ Tacit function, takes 1 argument (E.g. 5)

         ⍳ ⍝ Range 1 2 3 4 5

       !∘  ⍝ Factorials. Yields 1 2 6 24 120

    2*⍨    ⍝ Squared. Yields 1 4 36 576 14400

  !        ⍝ Factorial of the argument. Yields 120.

   ≤       ⍝ Less than or equal to. Yields 0 0 0 1 1

+/         ⍝ Sum the results, yielding 2.

Since this answer wasn't strictly made by me, I'll keep my original answer below.

APL (Dyalog Unicode), 14 12 bytes

+/(!>×)∘⌽∘⍳

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Prefix tacit function. Basically a Dyalog port of Jonathan's answer.

Thanks to ngn and H.PWiz for the help in chat.

Thanks to Dennis for pointing out that my original code was wrong. Turns out it saved me 2 bytes.

Uses ⎕IO←0.

How:

+/(!>×)∘⌽∘⍳ ⍝ Tacit function, taking 1 argument (E.g. 5).

           ⍳ ⍝ Range 0 1 2 3 4

         ⌽∘  ⍝ Then reverse, yielding 4 3 2 1 0

  (    )∘    ⍝ Compose with (or: "use as argument for")

   !         ⍝ Factorial (of each element in the vector), yielding 24 6 2 1 1

     ×      ⍝ Multiply scan. Yields 4 12 24 24 0

    >        ⍝ Is greater than. Yields 1 0 0 0 1

+/           ⍝ Finally, sum the result, yielding 2.

R, 70 41 38 bytes

-29 because Dennis knows all the internal functions

-3 switching to scan() input

sum(prod(x<-scan():1)<=cumprod(1:x)^2)

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Pretty simple R implementation of nedla2004's Python3 answer.

I feel like there's a cleaner implementation of the 1-handling, and I'd like to lose the curly-braces.

I'm mad I didn't go back to using a which approach, madder that I used a strict less-than, but even madder still that I didn't know there's a cumprod() function. Great optimisation by Dennis.

Haskell, 52 51 bytes

Using the straightforward approach: We check whether the product of the last $n$ numbers, which is $frac{x!}{(x-n)!}$ is less than the product of the first $n$ numbers, namely $(x-n)!$ and takes the least $n$ for which this is true.

g b=product[1..b]

f x=[n|n<-[1..],g(x-n)^2<=g x]!!0

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Clean, 57 bytes

import StdEnv

$x=while(e=prod[1..x-e]^2>prod[1..x])inc 1

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A straight-forward solution.

Jelly, 7 bytes

R!²<!ċ0

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How it works

R!²<!ċ0  Main link. Argument: n



R        Range; yield [1, ..., n].

 !       Map factorial over the range.

  ²      Take the squares of the factorials.

    !    Compute the factorial of n.

   <     Compare the squares with the factorial of n.

     ċ0  Count the number of zeroes.

Python 3, 183 176 149 bytes

R=reversed

def M(I,r=1):

 for i in I:r*=i;yield r

def f(x):S=[*range(1,x+1)];return([n for n,a,b in zip([0]+S,R([*M(S)]),[0,*M(R(S))])if b>a]+[x])[0]

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It's is a lot faster than some other solutions - 0(N) multiplications instead of O(N²) - but I can't manage to reduce code size.

-27 from Jo King

05AB1E, 15 11 bytes

E!IN-!n›iNq

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E!IN-!n›iNq



E                For loop with N from [1 ... input]

 !               Push factorial of input    

  IN-            Push input - N (x - n)

     !           Factorial

      n          Square

       ›         Push input! > (input - N)^2 or x! > (x - n)^2

        i        If, run code after if top of stack is 1 (found minimum number of candies)

         N       Push N

          q      Quit, and as nothing has been printed, N is implicitly printed

Uses the same approach as my Python submission. Very new to 05AB1E so any tips on code or explaination greatly appreciated.

-4 bytes thanks to Kevin Cruijssen

Jelly, 14 bytes

ạ‘rP>ạ!¥

1ç1#«

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Handles 1 correctly.

Charcoal, 20 bytes

ＮθＩ⊕ΣＥθ‹Π⊕…ιθ∨Π…¹⊕ι¹

Try it online! Link is to verbose version of code. Explanation:

Ｎθ                      Input `n`

    Σ                   Sum of

      θ                 `n`

     Ｅ                  Mapped over implicit range

        Π               Product of

           ι            Current value

          …             Range to

            θ           `n`

         ⊕              Incremented

       ‹                Less than

              Π         Product of

                ¹       Literal 1

               …        Range to

                  ι     Current value

                 ⊕      Incremented

             ∨          Logical Or

                   ¹    Literal 1

   ⊕                    Incremented

  Ｉ                     Cast to string

                        Implicitly print

Product on an empty list in Charcoal returns None rather than 1, so I have to logically Or it.

PHP, 107 bytes

<?php $x=fgets(STDIN);function f($i){return $i==0?:$i*f($i-1);}$n=1;while(f($x)<f($x-$n)**2){$n++;}echo $n;

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Uses the same $x^2>((x-1)!)^2$ method as others have used.

Uses the factorial function from the PHP submission for this challenge (thanks to @donutdan4114)

Wolfram Language (Mathematica), 43 bytes

Min[n/.Solve[#!>(#-n)!^2, Integers]]/.n->1&

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05AB1E, 7 bytes

L!ns!@O

Port of Dennis♦' Jelly answer, so make sure to upvote him if you like this answer!

Try it online or verify all test cases.

Explanation:

L          # List in the range [1, (implicit) input]

 !         # Take the factorial of each

  n        # Then square each

   s!      # Take the factorial of the input

     @     # Check for each value in the list if they are larger than or equal to the

           # input-faculty (1 if truthy; 0 if falsey)

      O    # Sum, so determine the amount of truthy checks (and output implicitly)

Japt -x, 7 bytes

Port of Dennis' Jelly solution.

Only works in practice up to n=4 as we get into scientific notation above that.

õÊ®²¨U²

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õ           :Range [1,input]

 Ê          :Factorial of each

  ®         :Map

   ²        :  Square

    ¨       :  Greater than or equal to

     U²     :  Input squared

            :Implicitly reduce by addition

C (gcc), 68 bytes

n;f(T x){T i=2,j=x,b=1,g=x;while(i<j)b*i>g?g*=--j:(b*=i++);n=x-j+1;}

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Edit: shove a few obvious bytes

Edit 2: trading bytes against mults, no doing 2*x mults instead of x+n

Well I have this in C. Fails at 34 when T is long.

There is a possible ambiguity as to whether the good kid wants to always win or never lose... what do you think?

C# (.NET Core), 93 bytes

n=>{int d(int k)=>k<2?1:k*d(k-1);int a=1,b=d(n),c=n;for(;;){a*=n;b/=n--;if(a>=b)return c-n;}}

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Based off of @Arnauld's javascript answer.