Wilson's Theorem/Necessary Condition/Proof 2

Proof
If $p = 2$ the result is obvious.

Therefore we assume that $p$ is an odd prime.

Consider $p$ points on the circumference of a circle $C$ dividing it into $p$ equal arcs.

By joining these points in a cycle, we can create polygons, some of them stellated.

From Number of Different n-gons that can be Inscribed in Circle, the number of different such polygons is $\dfrac {\paren {p - 1}!} 2$.

When you rotate these polygons through an angle of $\dfrac {2 \pi} p$, exactly $\dfrac {p - 1} 2$ are unaltered.

These are the regular $p$-gons and regular stellated $p$-gons.

That there are $\dfrac {p - 1} 2$ of them follows from Number of Regular Stellated Odd n-gons.

The remaining $\dfrac {\paren {p - 1}!} 2 - \dfrac {p - 1} 2$ polygons can be partitioned into sets of $p$ elements: those which can be obtained from each other by rotation through multiples of $\dfrac {2 \pi} p$.

The total number of such sets is then:


 * $\dfrac {\paren {p - 1}! - \paren {p - 1} } {2 p}$

Thus we have that:
 * $2 p \divides \paren {\paren {p - 1}! - \paren {p - 1} }$

That is:
 * $p \divides \paren {\paren {p - 1}! + 1}$