[CmKaN]

Pom dai only 5 please check
Case 1: if $p=5$ --> $5^{p^{2}+1} \equiv 0 \pmod{p^{2}}$
Case 2: if $p \neq 5$ --> $(5,p^{2})=1$
$\phi {p^{2}} = p-1$
$\therefore 5^{p-1} \equiv 1 \pmod{p^{2}}$
$5^{p^{2}+1} \equiv 5^{p+2} \equiv 0 \pmod{p^{2}}$
$\therefore p | 5$ --><--
$\therefore p=5$

14. $n \in \mathbb{N},n>1$ prove that $((n-1)!+1,n)=n$ <-->$n \in \mathbb{P}$