Prove that $(a-b) \mid (a^n-b^n)$ [duplicate]

I'm trying to prove by induction that for all $a,b \in \mathbb{Z}$ and $n \in \mathbb{N}$, that $(a-b) \mid (a^n-b^n)$. The base case was trivial, so I started by assuming that $(a-b) \mid (a^n-b^n)$. But I found that: \begin{align*} (a-b)(a^{n-1}+a^{n-2}b + a^{n-3}b^2+...+b^{n-1}) &= a^n-b^n. \end{align*}

Doesn't this imply that $(a-b) \mid (a^n-b^n)$ as $a^{n-1}+a^{n-2}b+\dots+b^{n-1}$ is clearly an integer? This obviously isn't a proof by induction, but is there anything wrong with taking this approach to prove this result, other than the fact that it isn't what is being asked?

Aside, I'm having trouble getting started with the proof by induction. I've tried making the above equivalence useful during the induction step, but it doesn't seem to be helpful, so any hints would be greatly appreciated!

The induction step can be handled by observing that $a^{(k+1)} - b^{(k+1)} = aa^k - ab^k + ab^k - bb^k = a(a^k - b^k) + (a - b)b^k$. Then by the inductive hypothesis, $a - b$ divides each summand.

Would say more but I must crash. Perhaps I can add the details of the full inductive proof manana. But it should be easy from here . . .

G'night, fellow math-heads!

Observe:

$a - b \equiv 0 \pmod {a-b} \implies a\equiv b \pmod{a-b} \implies a^n \equiv b^n \pmod{a-b} $