I am looking for a combinatorial proof of Wilson's Theorem. Something along the lines of this kind of proof.
$\textbf{Combinatorial proof of Fermat's Little Theorem}$
First consider a $p$ -tuple and let us say there are $a$ distinct numbers where $a > 0$ . Now the total possible tuples are $a^p$. Now we say $2$ tuples are equivalent if one is obtained by a cyclic permutation of the other. Now there are $a$ equivalence classes consisting of $1$ element and all other equivalence classes contain $p$ elements(Here we use the fact that $p$ is a prime). Now this implies that $a^{p-1} \equiv 1 \mathrm{mod} p$
There is a proof in Andrews book in number theory that goes as follows:
Consider a circunference with $p$ points that correspond to the vertices of a regular $p$-gon. How many polygons can have those vertices? $\frac{(p-1)!}{2}$. There are two types of polygons, those which are invariant under rotations of $\frac{2\pi}{p}$ radians, and those which give $p$ different polygons when rotated $p$ times.
How many are there of the first type? they are those which follow the rule that vertex with number $n$ is connected to vertex numbered $n+k$. There are $\frac{p-1}{2}$ of these.
We conclude $\frac{(p-1)!}{2}-\frac{p-1}{2}$ is a multiple of $p$. thus $(p-1)!-(p-1)$ is a multiple of $p$. so $2(p-1)!+1$ is a multiple of $2$, so $(p-1)!\equiv-1\bmod p$.