Foundational proof for Mersenne primes

Question

I know how to prove that, if $2^n-1$ is prime and $n>1$, then $n$ is prime.

But how do we prove that, if $a^n-1$ is prime and $n>1$, then $a$ must equal 2?

Answer 1

If we say $P(a)=a^n-1$ we have that $P(1)=0$ and by Polynomial remainder theorem we get that $$a-1\mid a^n-1$$