If $n^k$ is odd, $n$ is odd (proof by contra-positive) I started with the claim that: If $n$ is even, $n^k$ is even.
But I have only gotten this far:
$$n = 2a,$$
$$n^k = (2a)^k,$$
$$n^k = 2^k a^k.$$
I am trying to reach the format where I can show that
$$n^k = 2a,$$
so that $n^k$ is even.

I will use your solution. Let's assume that $n$ is even, then we can write
$$n=2a.$$
Powering the both sides we obtain
$$n^k=(2a)^k=2^ka^k.$$
Taking $a'=2^{k-1}a^k$ we have
$$n^k=(2a)^k=2^ka^k=2\cdot 2^{k-1}a^k=2a',$$
so $n^k$ is even.
