M.I. Divisibility problems

In your mathematics textbook on Mathematical Induction, you should have seen many problems like these:

Prove by M.I. that
$9^n-4^n$ is divisible by 5.
$5^n-1$ is divisible by 4.
$7^n-5^n$ is divisible by 2, etc.

Now do you notice something? It seems that $a-b \vert a^n-b^n$, that is, $a-b$ divides $a^n-b^n$.

In fact, you're right. There is an identity: $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1})$
(Expand RHS to see they are equal.)

More interestingly, this is related to modular arithmetic.
$a \equiv b$ (mod a-b)
$a^n \equiv b^n$ (mod a-b)
$a^n-b^n \equiv 0$ (mod a-b)
This implies that $a^n-b^n$ is divisible by a-b.