Prove that $\big( n^2 - 1 \big) \mid \big( 2^{n!} - 1 \big) $

Question

Here's a problem from 104 Number Theory Problems.

Prove that for any even positive integer $n$, $n^2-1$ divides $2^{n!} - 1$.

Solution. Let $m=n+1$. We need to prove that $m(m-2)$ divides $2^{(m-1)!}-1$. Because $\phi(m)$ divides $(m-1)!$ we have $\left(2^{\phi(m)}-1\right) \mid \left(2^{(m-1)!} - 1 \right)$ and from Euler's theorem, $ m \mid \left( 2^{\phi(m) - 1} \right) $, so it follows that $ m \mid \left( 2^{(m-1)!} - 1 \right) $. Similarly, $ \left( m - 2 \right) \mid \left( 2^{(m-1)!} - 1 \right) $. Because $m$ is odd, $ \text{gcd}(m,m-2) = 1 $ and the conclusion follows. $\Box$

Uh, excuse me, how does $ \phi(m) \mid (m-1)! $ imply $ \left( 2^{\phi(m)}-1 \right) \mid \left( 2^{(m-1)!} - 1 \right) $ ?

Answer 1

$$2^{ab}- 1 = (2^a -1)\left(\sum_{k=0}^{b-1}2^{ka}\right)$$

Answer 2

I find it easier to see it this way. $$ 2^a\equiv1\pmod {2^a−1} ⇒2^{ab}\equiv1^b\equiv1 \pmod{2^a−1} $$