Fermat's Little Theorem: Proving p-1! ≡ -1 (mod p)

[moonlight310]

Homework Statement

let p be prime then, (p-1)! is congruent to -1 mod p

Homework Equations

The Attempt at a Solution

I'm not sure where to start

 

[VeeEight]

First of all you should try some examples out for small prime.

Since p is prime, the set {1, 2, ... p-1} is a group under multiplication, modulo p. This means that there is a (unique) inverse for each element.

 

[moonlight310]

I've tried small numbers and it works. So since it has an inverse it means that it can be mod p ? I'm sorry I don't understand this stuff very well.

 

[VeeEight]

The fact there there is an inverse means that for each element x, there is an element y such that xy = 1 mod p - and this is only true because p is a prime (a well known group theory result). The main idea for the proof of this theorem is to try to pair up each element with it's inverse (which is valid since this group is commutative).