APL (Dyalog Unicode), 8 5 bytes

⍳*⊢-⍳

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Anonymous prefix tacit function. TIO tests for the range [1..10].

Thanks @lirtosiast for 3 bytes.

How:

⍳*⊢-⍳ ⍝ Tacit function

    ⍳ ⍝ Range. ⍳n generates the vector [1..n].

  ⊢- ⍝ Subtracted from the argument. The vector is now [n-1,n-2,...,0]

⍳*    ⍝ Exponentiate using the range [1..n] as base. The result is the vector

      ⍝ [1^(n-1), 2^(n-2), 3^(n-3),...]

Japt, 5 bytes

õ_p´U

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õ         :Range [1,input]

 _        :Map

  p       :  Raise to the power of

   ´U     :  Input decremented

Perl 6, 19 bytes

{^$_+1 Z**[R,] ^$_}

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Anonymous code block that takes a number and returns a list. Zip exponents the range 1 to input and the range input-1 to 0

Aheui (esotope), 193 164 bytes (56 chars)

방빠싹받분샥퍼붇바파쟈뿌차샦히망맣여

타빠바푸투반또분뽀뿌서썪삯타삯받반타

석차샦져쌲볼어타토싻삭빠쏛ㅇ또섞썪뻐

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Try it on AVIS(Korean); just copy and paste code above, press start button, input a number, see how it moves. To see output, press the >_ icon on left side.

It's not golfed much, but I give it a shot.

Jelly, 5 bytes

R*ḶU$

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R                [1,...,n]

 *               to the power of

  ḶU$            [0,...,n-1] reversed

Pyth, 5 bytes

_m^-Q

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Optimally encoded this would be 4.106 bytes.

_                reverse of the following list:

 m               map the following lambda d:

  ^                (N-d)**d

   -Qd             

      d

       Q         over [0,...,N-1]

Haskell, 23 bytes

f i=[x^(i-x)|x<-[1..i]]

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Alternative version, also 23 bytes:

f i=(^)<*>(i-)<$>[1..i]

J, 10 bytes

(>:^|.)@i.

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If we really need to separate the numbers by a newline:

J, 13 bytes

,.@(>:^|.)@i.

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PHP, 32 bytes

while($argn)echo++$i**--$argn,_;

Run as pipe with -nR or try it online.

Octave, 18 bytes

@(n)(t=1:n).^(n-t)

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Thanks Luis Mendo, using internal variable saves 3 bytes.

Wolfram Language (Mathematica), 24 20 18 bytes

(x=Range@#)^(#-x)&

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-4 thanks @lirtosiast.

MathGolf, 6 bytes

rx╒m#

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Python 2, 40 bytes

lambda n:[i**(n-i)for i in range(1,n+1)]   #Outputs a list

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Python 2, 41 bytes

n,i=input(),0

exec"print(n-i)**i;i+=1;"*n   #Prints in reversed order

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Ruby, 27 bytes

->n{(1..n).map{|r|r**n-=1}}

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Retina, 35 bytes

.+

*

_

$$.($.'*$($.>`$*)_¶

%~`^

.+¶

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.+

*

Convert the input to unary.

_

Match each position. This then sets several replacement variables. $` becomes the left of the match; $>` modifies this to be the left and match; $.>` modifies this to take the length, i.e. the current index. $' meanwhile is the right of the match, so $.' is the length i.e. the current exponent.

$$.($.'*$($.>`$*)_¶

Create a string $.( plus $.' repetitions of $.>`* plus _. For an example, for an index of 2 in an original input of 5, $.' is 3 and $.>` is 2 so the resulting string is $.(2*2*2*_. This conveniently is a Retina replacement expression that caluclates 2³. Each string is output on its own line.

%~`^

.+¶

For each line generated by the previous stage, prefix a line .+ to it, turning it into a replacement stage, and evaluate that stage, thereby calculating the expression.

QBasic, 3533 bytes

Thank you @Neil for 2 bytes!

INPUT a

FOR b=1TO a

?b^(a-b)

NEXT

Slightly expanded version on REPL.IT because the interpreter in't entirely up-to-spec.

Output

QBasic (qb.js)

Copyright (c) 2010 Steve Hanov



   5

1

8

9

4

1

F# (.NET Core), 42 bytes

let f x=Seq.map(fun y->pown y (x-y))[1..x]

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JavaScript (Node.js), 33 32 bytes

n=>(g=i=>--n?++i**n+[,g(i)]:1)``

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-3 bytes with credits to @Shaggy, and -1 byte by @l4m2!

JavaScript (Node.js), 36 bytes

f=(n,i=1)=>n--?[i++**n,...f(n,i)]:

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JavaScript (Node.js), 37 bytes

n=>[...Array(n)].map(x=>++i**--n,i=0)

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C# (Visual C# Interactive Compiler), 46 bytes

x=>new int[x].Select((_,i)=>Math.Pow(i+1,--x))

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MATL, 5 bytes

:Gy-^

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Explanation

Consider input 5 as an example.

:     % Implicit input. Range

      % STACK: [1 2 3 4 5]

G     % Push input again

      % STACK: [1 2 3 4 5], 5

y     % Duplicate from below

      % STACK: [1 2 3 4 5], 5, [1 2 3 4 5]

-     % Subtract, element-wise

      % STACK: [1 2 3 4 5], [4 3 2 1 0]

^     % Power, element-wise. Implicit display

      % STACK: [1 8 9 4 1]

Java, 59 Bytes

for(int i=1;a+1>i;i++)System.out.println(Math.pow(i,a-i));

Jelly, 4 bytes

*ạ¥€

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Clean, 37 bytes

import StdEnv

$n=[i^(n-i)\i<-[1..n]]

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Defines $ :: Int -> [Int] taking an integer and returning the list of results.

$ n                // function $ of n

 = [i ^ (n-i)      // i to the power of n minus i

    \ i <- [1..n] // for each i in 1 to n

   ]

R, 34 bytes

x=1:scan();cat(x^rev(x-1),sep=',')

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05AB1E, 5 bytes

LD<Rm

Port of @lirtosiast's Jelly answer.

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Explanation:

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

       #  i.e. 5 → [1,2,3,4,5]

 D<    # Duplicate this list, and subtract 1 to make the range [0, input)

       #  i.e. [1,2,3,4,5] → [0,1,2,3,4]

   R   # Reverse it to make the range (input, 0]

       #  i.e. [0,1,2,3,4] → [4,3,2,1,0]

    m  # Take the power of the numbers in the lists (at the same indices)

       # (and output implicitly)

       #  i.e. [1,2,3,4,5] and [4,3,2,1,0] → [1,8,9,4,1]

Lua, 43 41 bytes

-2 bytes thanks to @Shaggy

s=io.read()for i=1,s do print(i^(s-i))end

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R, 22 bytes

n=scan();(1:n)^(n:1-1)

Fairly self-explanatory; note that the : operator is higher precendence than the - operator so that n:1-1 is shorter than (n-1):0

If we are allowed to start at 0, then we can lose two bytes by using (0:n)^(n:0) avoiding the need for a -1.

Charcoal, 9 bytes

Ｉ⮌ＥＮＸ⁻θιι

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   Ｎ        Input as a number

  Ｅ         Map over implicit range

       ι    Current value

     ⁻      Subtracted from

      θ     First input

    Ｘ       Raised to power

        ι   Current value

 ⮌          Reverse list

Ｉ           Cast to string

             Implicitly print on separate lines

C# (Visual C# Interactive Compiler), 55 bytes

v=>Enumerable.Range(0,v--).Select(i=>Math.Pow(i+1,v--))

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Perl 5 -n, 21 bytes

say++$**--$_ while$_

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