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.

$$n=2m\iff n^k=2p$$ where $p=2^{k-1}m^k$.
