Prove that $\forall n > 1 \quad2^n - 1 \pmod n \neq 0$ Prove that $\forall n > 1, \quad2^n - 1 \pmod n \neq 0$
I've thought of the induction but I can't figure out how to prove the step. Fermat's theorem (and its variations) aren't particularly useful as well. Any hint would be greatly appreciated!

Assume you have got a $n>1$, so that $2^n \equiv 1 \pmod{n}$. Take the smallest prime divisor $p$ of $n$.
You have $p\;|\;n\;|\;2^n-1$, hence $2^n\equiv 1 \pmod{p}$.
But you have also, from Fermat's little theorem, that $2^{p-1}\equiv 1\pmod{p}$.
Since $p-1$ and $n$ are coprime, you have thus by Bézout's identity, some $a,b$ such that $a(p-1)+bn=1$, and then
$$1\equiv(2^{p-1})^a\cdot (2^n)^b \equiv 2^{a(p-1)+bn}\equiv 2^1 \pmod{p}$$
Thus $2^1\equiv 1\pmod{p}$, or equivalently $p\;|\;1$, but this is not possible.
