How do I show this :for odd $n$ :$\displaystyle\ (2^{n!}-1) \bmod n=0$?

Question

I have tried to show this using fermat little theorem but i can't succed: for odd $n$ , $2^{n!}-1\equiv 0\pmod n$. I have used the factorisation of $n!$ with $n$ is odd integer greater than $1$ as $ p_{1}^{e_{1}}\cdots p_{k}^{e_{k}}$ for using Fermat's little theorem but it's hard for me ...?

Then, how do I show that for odd $n$: $2^{n!}-1\equiv 0 \pmod n$?

Thank you for any help

Answer 1

By assumption, $\gcd (2,n)=1$. Therefore $2^{\varphi(n)}\equiv 1 \pmod n$. But of course $$\varphi(n)<n\implies \varphi(n)\,|\,n!$$