group theoretic proof of Wilson’s theorem

Here we present a group theoretic proof of it.
Clearly, it is enough to show that$(p-2)!\equiv 1(modp)$ since $p-1\equiv -1(modp)$.By Sylow theorems, we have that $p$-Sylow subgroups of $S_p$, thesymmetric group on $p$ elements, have order $p$, and the number $n_p$ ofSylow subgroups is congruent to 1 modulo $p$. Let $P$ be a Sylow subgroupof $S_p$. Note that $P$ is generated by a $p$-cycle. There are $(p-1)!$ cyclesof length $p$ in $S_p$. Each $p$-Sylow subgroup contains $p-1$ cyclesof length $p$, hence there are $\frac{(p-1)!}{p-1}=(p-2)!$ different$p$-Sylow subgrups in $S_p$, i.e. $n_P=(p-2)!$. From Sylow’s SecondTheorem, it follows that $(p-2)!\equiv 1(modp)$,so $(p-1)!\equiv -1(modp)$.