TITLE: Prove that $\forall n > 1 \quad2^n - 1 \pmod n \neq 0$
QUESTION [5 upvotes]: 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!

REPLY [4 votes]: A counterexample $\,n\,$ i.e.  $\ n\mid 2^n-1\,$ would contradicts the more general theorem below.
Theorem $\,\ \ \ m,n>1,\,\ m\mid 2^{\large n}-1\,\Rightarrow\, \ell(m) > \ell(n),\ \ $   $\ell(x) =\,$ least prime factor of $\,x $ 
Proof $\quad\ \, {\rm mod}\,\ p = \ell(m)\!:\,\ 2^{\large n}\equiv 1\equiv\, 2^{\large\color{}{ p-1}}\,$ $\Rightarrow$ $\,\ \ell(m) = p > {\rm ord}\,2\color{#80f}{ \ge \ell(n)}$ 
Remark $\,\ $ The $ $ key dea $ $ is: $\ \  2^n\equiv 1,\,\ \color{#0a0}{2\not\equiv 1}\,$ implies the order of $\,2\,$ is $\,\color{#80f}{\ge\ {\rm least\  prime}\,\ q\mid n},\ $ because the order  must divide $\,n,\,$ and is $\color{}{\neq \color{#c00}1}$ (else $\,\color{#0a0}{2^{\color{#c00}{\large 1}}\equiv 1})$.