1. ## compostive

n!+1 is composite.

2. Hint given by book: use Wilson's theorem to guarantee that n!+1 has a prime factor for an appropriately chosen n

3. Let $n=p-1$ then $n! + 1 = (p-1)! + 1 \equiv 0 (\bmod p)$.
Thus, $p$ is a prime divisor.

4. Originally Posted by ThePerfectHacker
Let $n=p-1$ then $n! + 1 = (p-1)! + 1 \equiv 0 (\bmod p)$.
Thus, $p$ is a prime divisor.
would n! -1 be the same thing setting n=p-1 and getting (p-1)!-1 but since 1 is the only inverse modulo of itself it is the same thing as (p-1)!+1??