Here we present a group theoretic proof of it.
Clearly, it is enough to show that $ (p-2)!\equiv 1\pmod{p}$ since $ p-1\equiv -1\pmod{p}$ . By Sylow theorems, we have that $p$ -Sylow subgroups of $S_p$ , the symmetric group on $p$ elements, have order $p$ , and the number $n_p$ of Sylow subgroups is congruent to 1 modulo $p$ . Let $P$ be a Sylow subgroup of $S_p$ . Note that $P$ is generated by a $p$ -cycle. There are $(p-1)!$ cycles of length $p$ in $S_p$ . Each $p$ -Sylow subgroup contains $p-1$ cycles of 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 Second Theorem, it follows that $(p-2)!\equiv1\pmod{p}$ ,so $(p-1)!\equiv-1\pmod{p}$ .