# Prove that for every even positive integer $n$, $n^2 − 1$ divides $2^{n!} − 1$.

I am trying to prove that for every even positive integer $$n$$, $$n^2 − 1$$ divides $$2^{n!} − 1$$.

My attempt: I am thinking of using Euler's Theorem and totient function to get $$2^{n!} \equiv 1$$ (mod $$n^2 - 1$$). We would have to show $$\text{gcd}(2^{n!} - 1, n^2 − 1) = 1$$ however and I'm not sure how to proceed with this.

Note that $$2^{n!}\equiv1\bmod{n+1}$$ and $$2^{n!}\equiv1\bmod{n-1}$$
This is because $$\phi(n+1),\phi(n-1), so $$\phi(n+1),\phi(n-1)\mid n!$$, and you know that $$2^{\phi(n-1)}\equiv1\bmod{n-1}$$ by Euler's Theorem (and similarly for $$n+1$$).
Since $$\operatorname{gcd}(n+1,n-1)=1$$ since $$n$$ is even, by CRT, $$2^{n!}\equiv1\bmod{n^2-1}$$. So, $$n^2-1\mid2^{n!}-1$$