(a) Prove that if n > 2, then there exists a prime p satisfying n < p < n!.
[Hint: If n! − 1 is not prime, then it has a prime divisor p; and p ≤ n implies p|n!, leading to a contradiction.]
(b) For n > 1, show that every prime divisor of n! + 1 is an odd integer that is greater than n.
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