So this is my first challenge on this site.

What this challenge is is that you input a number, and your code spits out numbers which can be calculated as an exponent. The sum of the exponent and base in the number must equal the input.
The numbers must start from $1^{n-1}$ to $(n)^0$.

Example

Given input 5, the program will print:

1  

8  

9  

4  

1  

$1^4$ is 1 and $1+4=5$
$2^3$ is 8 and $2+3=5$ and so on

Input and Output

Input will be in the form of an integer.
Output will be a list of numbers, delimited by either commas or new lines.

This is code-golf, so shortest code wins.

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

   ]

Wolfram Language (Mathematica), 24 20 18 bytes

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

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-4 thanks @lirtosiast.

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]

MathGolf, 6 bytes

rx╒m#

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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]

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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Jelly, 5 bytes

R*ḶU$

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R                [1,...,n]

 *               to the power of

  ḶU$            [0,...,n-1] reversed