TITLE: Prove that for any even positive integer $n$, $n^2-1 \mid 2^{n!}-1$
QUESTION [7 upvotes]: Prove that for any even positive integer $n$, $n^2-1 \mid 2^{n!}-1$  

This is from a book. They have given the proof. But I didn't understand it well. I am looking for a simpler proof. Or it will be helpful if someone explain this a little bit more -
Proof: Let $m = n+1$ then we need to prove that $m(m-2) \mid 2^{(m-1)!}-1$. Because of $\phi(m) \mid (m-1)!$, we have $2^{\phi(m)} -1 \mid 2^{(m-1)!} -1$. And from Euler's theorem, $(m-2) \mid 2^{(m-1)!}-1$. Because $m$ is odd, $gcd(m,m-2)=1$ and the conclusion follows.

REPLY [2 votes]: $2\,$ is coprime to $\,n\!+\!1\,$ so $\,{\rm mod}\ n\!+\!1\!:\, $ $\,2$ has order $\le n,\,$  i.e. $\,\color{#c00}{2^{\large k}\! \equiv 1}\,$ for $\,k\le n,\,$ thus $\,k\mid n!\:$ hence $\,2^{\large n!}\!\equiv (\color{#c00}{2^{\large k}})^{\large n!/k}\!\equiv 1.\,$  Similarly $\,2^{\large n!}\!\equiv 1\pmod{\!n\!-\!1}.$ Thus $\,2^{\large n!}\!-1\,$ is divisible by $\,n\!+\!1\,$ and $\,n\!-\!1\,$ hence also by their lcm = product, since $\:\gcd(n\!+\!1,n\!-\!1) = \gcd(n\!+\!1,2) = 1,\,$ by $n$ even.

REPLY [2 votes]: I have a shorter proof, because $n$ is even then $n^2-1$ is odd and $\gcd(n^2-1,2)=1$, thus according to Euler's theorem
$$2^{\varphi(n^2-1)}\equiv 1 \pmod{n^2-1}$$
But totient function is multiplicative and $\gcd(n-1,n+1)=1$ or
$$\varphi(n^2-1)=\varphi(n+1)\cdot \varphi(n-1)\leq n\cdot (n-2)<n!$$
or $n!=\varphi(n^2-1)\cdot Q, Q \in \mathbb{N}$ and
$$2^{n!} \equiv 2^{\varphi(n^2-1)\cdot Q} \equiv 1^{Q} \equiv 1 \pmod{n^2-1}$$