How can I prove that if $\displaystyle a^n - 1$ is prime, then $\displaystyle a = 2$ and $\displaystyle n$ is prime?
I didn't quite follow your explanation there. My explanation is
$\displaystyle a \equiv 1\ (\text{mod}\ (a-1))$
Therefore $\displaystyle a^n-1 \equiv 1^n - 1 \equiv 0\ (\text{mod}\ (a-1))$.
Edit: Never mind, I guess you weren't trying to explain that part. The wording is just a bit odd though, in terms of, we haven't proven that $\displaystyle a^n-1$ is prime, merely stating conditions for it being possible.