Exercise 1.3.44 (Mersenne numbers.)

Proof. If $n=1$, $2^n-1=1$ is not prime.

If $n$ is composite, then $n=\mathit{qr}$, where $1<q<n$. Then

where $1=2^1-1<2^q-1<2^n-1=N$. Therefore $2^q-1$ is a non trivial divisor of $N$, so $N=2^n-1$ is composite.

Hence, If $2^n-1$ is prime for some integer $n$, $n$ is itself prime. □