modular arithmetic - Prove by induction that if $aequiv b pmod m$ then $a^n equiv b^n pmod m$

The base case is pretty straightforward. But I'm stuck on the inductive step.

As the base case holds, assume for when $n=k$ holds, show the $k+1$ case holds true.

Inductive Hypothesis: $a^k \equiv b^k \pmod m$, then

$$a^{k+1} \equiv b^{k+1} \pmod m \iff a^{k+1} - b^{k+1} = m(k), k \in \mathbb{Z}.$$

I think I'm missing some steps, I'm not sure how to manipulate what I have to shows it holds.