Prove that for any $n > 0$, if $a^n$ is even, then $a$ is even the proof at hand is:

Prove that for any $n > 0$, if $a^n$ is even, then $a$ is even.
Hint: Contradiction

So I know to start the problem in a contradiction format would be:
$a^n$ is even, then $a$ is odd so that $a = 2k+1$.  Then plug in that into $a^n$, we get $(2k+1)^n$.  This is where I become stuck.
Thanks in advance!

You can prove this by induction: 
For $n=1$ the claim is clear. 
Assume that if $a^n $ is even then $a $ is even. 
Let $a^{n+1}$ be even. Then:
$a^{n+1} = a a^n$ .
If a is not even, then $a^n$ must be even because otherwise their product won't be even. 
But then , by the induction hypothesis , $a$ is even, contradiction. 
So $a$ must be even, as wanted.
