how to find all integer $n$ such that $ n\mid2^{n!}-1$
I find:
Of course $2 \nmid n$. We prove that, if $2 \nmid n$ then $n \mid 2^{n!}-1$. $2 \nmid n \iff n = 2k+1 , k \ge 0$, we'll prove:
$2^{(2k+1)!} \equiv 1\pmod{2k+1}$
Let $n = p_1^{a_1}\cdot p_2^{a_2} \cdot ... \cdot p_s^{a_s}$, we'll prove that:
$2^{(2k+1)!} \equiv 1 \pmod{p_1^{a_1}}$
Let $t = ord_{p_1^{a_1}}2 \iff 2^t \equiv 1\pmod{p_1^{a_1}}$, so:
$t \mid (2k+1)! \iff (2k+1)! = l\cdot t \iff l = \frac{(2k+1)!}{t} \in \mathbb{Z}_{+}$
And $2^t \equiv 1\pmod{p_1^{a_1}}/^l \Rightarrow 2^{(2k+1)!} \equiv 1\pmod{p_1^{a_1}}$
Analogously we show divisibility $2^{(2k+1)!}-1$ by $p_2^{a_2} , ... , p_s^{a_s}$