Let n be an odd positive number. Prove that n | $(2^{n!}-1)$.

Question

Let n be an odd positive number. Prove that n | $(2^{n!}-1)$. I don't know how to start this. If n is prime I might try Fermat's Little Theorem or something but as n is merely odd I don't know what to do.

Answer 1

Hint Let $\phi(n)$ be the Euler totient function. Since $\phi(n) \leq n$ we have $\phi(n)|n!$.