341Using Wilson's Theorem we have
(p-1)! \equiv -1 \pmod p \Rightarrow (p-2)! (p-1) \equiv -1 \pmod p
But (p-1) \equiv -1 \pmod p and hence
(p-2)! (-1) \equiv -1 \pmod p \Rightarrow (p-2)! \equiv 1 \pmod p
From this we deduce that
(p-2)! =(p-2r)! (p-2r+1)(p-2r+2)...(p-2) \equiv 1 \pmod p
Just as above we use that (p-k) =-k \pmod p
Thus we get
(p-2)! \equiv (p-2r)! (2r-1)(2r-2)...2 \equiv (p-2r)! (2r-1)! \equiv 1 \pmod p
In other words p|(p-2r)! (2r-1)! - 1
