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!

The contrapositive is that $$a \text{ odd }\implies a^n \text { odd}$$
Use induction here: if $a^1=2k_1+1$, then $a^2=(2k_1+1)^2=2(2k_1^2+2k_1)+1=2k_2+1$
hence true for $n=2$
Then assume true for $n=m$, $(2k_1+1)^m=2k_m+1$
Then $$(2k_1+1)^{m+1}=(2k_m+1)(2k_1+1)=4k_mk_1+2k_m+2k_1+1=2(2k_mk_1+k_m+k_1)+1$$
Hence true with $k_{m+1}=2k_mk_1+k_m+k_1$
