Prove that each prime divisor of $(2^p)-1$, where $p$ is a prime, is greater than $p$.

Question

How would I prove that if $(2^p)-1$ is not prime, there is not a prime number less than p that divides it? (any hints for a proof by contradiction?) I have assumed that there is a prime divisor less than p, let's say q. But I am unsure of where to go from there.

Answer 1

By Fermat, we know $q\mid 2^{q-1}-1$. Than note that $q\mid 2^a-1$ and $q\mid 2^b-1$ implies $q\mid 2^{\gcd(a,b)}-1$.