Here is a proof of Wilson's theorem, that $(p-1)!\equiv -1\,\text{mod}\,p$.

Sylow's 3rd theorem implies that the symmetric group $S_p$ (order $p!$) has $1\,\text{mod}\,p$ subgroups $\mathbb{Z}/p\mathbb{Z}$ cyclic, and since they all only intersect at the 0 element ($p$ is prime) there are a total of $(p-1)\cdot1\,\text{mod}\,p\equiv-1\,\text{mod}\,p$ elements of order $p$, which equivalently are the $(p-1)!$ amount of $p$-cycles.