Prove that if n4 is Not prime, then n divides (n-1)!, that is, (n-1)! is congruent to 0(mod n)

Question

Prove that if n>4 is Not prime, then n divides (n-1)!, that is, (n-1)! is congruent to 0(mod n)

anonymous · Answer

(n-1)! (a.k.a gamma(n))$$\Gamma(n)$$
equals to $$\prod_{i=1}^{n-1} i$$
in the case of n>4 and n in the set of primes;
There must exist some numbers (p(1),p(2),p(3),... p(N)) from all numbers smaller than n such that their total product is equal to n;
in $$\prod_{i=1}^{n-1} i$$
all the numbers from (p1,p2,p3,... pN) are of its components
therefore the product equals to $$\prod_{i=1}^{N} p(i)$$ (which is equal to n) times some integer value.