The proof is:
Prove that the value of (p-1)! mod p is always p-1.
Not really sure how to approach this so any help would be greatly appreciated!
Thanks! So basically the proof is this?:
In order to compute (p-1)!, you have to find the product of the integers 1 through p-1. In this set of integers, there are groups of two elements (excluding the group 1 and p-1) that can be multiplied together to get a number (p-1)!/(p-1), which equals p-1, which is congruent to 1 mod p.