Show that $n$ | $2^{n!}-1$ for all odd $n$.
I tried to show this by using Fermat's theorem but was unable to reach anything close. Since $n!$ is even then $2^{n!}$ is a square and so $2^{n!}$$-1$ is a difference of squares. Now we can assume that ${n!}$ = $2k$ and factorize it as ($2^k$$-1$)($2^k$$+1$). So we want to show that $n$ divides any one of these 2 factors. How exactly can we reach that?