Wilson's Theorem: (p-1)! = -1 (mod p) if and only if p is a prime.

1) It's what's "expected" to be used to answer this.

The use is that, if p is a prime, then -1 = (p-1)! = (p-1)(p-2)! = -(p-2)! (mod p), so that (p-2)! = 1 (mod p), so p divides [(p-2)!-1].

Ask yourself which k could make this fraction an integer: \(\displaystyle \frac{(p-1)! - (p-1)}{k} = \frac{(p-1) \ [(p-2)! - 1]}{k}\)

2) The other approach to this problem doesn't involve a real mathematical understanding, but rather understanding a trick for how to answer math questions on a test. So, this isn't the ideal way to do it, but it is a way, given that you trust that your teacher gave you a "correct" problem that didn't have an error.

When you're given a multiple choice question "for any prime p", then the answer has to be the same "for any prime p". In particular, the answer has to work for the prime p = 5. Only one of those choies, (a)-(e), holds when p=5, and so that choice must be the correct answer.