Prove that if $\displaystyle a^n -1 $ is prime where $\displaystyle a \in N $ and $\displaystyle n \geq 2, $ then $\displaystyle a =2 $ and $\displaystyle n $ is prime

How I can use the identity : $\displaystyle a^{kl} -1 $=$\displaystyle (a^k -1)(a^{k(l-1)} + a^{k(l-2)}$ + …+$\displaystyle a^{k }+1) $ to solve this Q ??