$2^p -1$ then $p$ is a prime

Question

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.

Answer 1

If $p$ is not a prime then $p=ab$ where $a\neq1,b\neq1$. Then $2^p-1=2^{ab}-1$. Now as $a-b\mid a^n-b^n$, hence $2^a-1\mid2^p-1$. $2^a-1\neq1$ as $a\neq1$ and $2^a-1\neq2^p-1$ as $b\neq1$. Hence $2^p-1$ is not a prime. Contradiction.

Answer 2

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