$2^p -1$ then $p$ is a prime I have been trying to solve this following problem:
If $2^p-1$ is a prime then prove that $p$ is a prime, where $p \geq 2$. 
which way should I go to prove this? Using fermat's or Bezout's theorem? or is it something else? 
This is what I managed to come up with,I am confused about it and  not sure if it is correct:
If $p$ is a prime and $2 \nmid p \rightarrow$ gcd$(2,p)=1$
so,  $2^{p-1} \equiv 1(p) $
$2^p\equiv2(p)$ 
so, $2^p\not\equiv 1(p) $ and since $2\not\equiv 1 (p)$ 
conc: $p\nmid2^p -1$ 
Any hints are appreciated.

If $p=mn$, where $m>1$ and $m>1$ then $2^{mn}-1$ is divided by $2^m-1$ and by $2^{n}-1$.
