Divisibility of Mersenne numbers Is there a way to prove that $2$ is the only prime that never divides $2^n-1$ ? Obviously we can ignore all primes that are themselves of this form. Some other examples:
$$5\,|\,2^4-1 \qquad 9\,|\,2^6-1 \qquad 11\,|\,2^{10}-1 \qquad 13\,|\,2^{12}-1 \qquad 17\,|\,2^8-1$$
I checked for all $p<10^6$. Thanks in advance.

See Fermat's little theorem:
https://en.wikipedia.org/wiki/Fermat%27s_little_theorem
Your claim follows with $n=p-1$.
