$n = 2^p - 1$ Prove that if n is prime then p must also be prime. My first thought was to try a contradiction; So given n is prime assume p is not prime i.e $p = p_{1}^{\alpha1} .... p_{r}^{\alpha r}$. But i didnt know where to go from there.

If $p = a\cdot b $ where $a \neq 1, b \neq 1$ then $n = 2^{ab} - 1 = (2^a)^b - 1 = (2^a-1)(2^{a(b-1)} + ...+ 1)$ is clearly composite.
