Problem of Chapter 15. A Consice Introduction to Pure Mathmatics

Given that p is a prime, prove that (p-1)! mod p = -1

In the following, a = b means a = b mod p

S = {Integer >= 2 and <= p-2}

1) a belongs to S ==> a * a not equals to 1 mod p (if a*a = 1, then a = +/- 1)
2) a belongs to S ==> a*b = 1 and b belongs to s ( p not divides a)
3) a belongs to S, a*b = 1 and a*c = 1, and b, c belongs to S ==> b = c
i.e. we have an unique partner a' in S for a, so that a*a' = 1

There are p-3 integers in S. According to above, we can group those integers into (p-3)/2 pairs, so that each pair's product is 1. Therefore

(p-2)x(p-3) ...3 x 2 = 1

==>

(p-1)! = (p-1) x (p-2)x(p-3) ...3 x 2 = -1