Prove that is p (greater than or equal to) 5 is a prime then (p^2) is congruent to 1(mod24).

I have a hint that a positive integer k divides on of k sequential integers n,(n+1),(n+2),...,(n+k-1), where n is an integer. the product of any k sequential integers n(n+1)(n+2)...(n+k-1) is congruent to 0(mod k)