A simple thing can be the following to prove that (q+1) divides q! if (q+1) is composite. Write q!=(q+1)!/(q+1)... Now we have to prove that (q+1)! is divisible by (q+1)² if (q+1) is composite. Decompose (q+1) to m×n where m and n are two integers (which are less than (q+1)). Factors of (q+1) are of course less than (q+1) and since (q+1)! contains all integers 1≤k≤(q+1) then m,n will lie too. Hence m,n and (q+1) being present in (q+1)! proves that (q+1)! is divisible by (q+1)² and thus q! is divisible by (q+1) if it is composite.
@@henryginn7490 I really appreciate that! It's a nice argument. When I say (n-1)!-1 I was immediately drawn towards Wilson's theorem and if that had any relation, and it sort of does!
