An abstract algebraic proof that if for some positive integer $n$, $2^n - 1$ is prime, then so is $n$. I was wondering if there exists a proof from abstract algebra of this theorem. It seems like there should be. In fact, I've tried to come up with one and it always seems I'm coming close, but I can't do it.
Could you provide one?

If $n$ is not prime, then $2^n - 1$ is not prime.
Suppose that $n = ab$ for integers $a,b\geq2$.  Then note that we can factor
$$
x^n - 1 = (x^a)^b - 1 = (x^a - 1)([\text{stuff}])
$$
which means that
$$
2^n - 1 = (2^a)^b - 1 = (2^a - 1)([\text{stuff}])
$$
where stuff is, notably, an integer and bigger than $1$. It follows that $2^n-1$ is not prime.
