Exercise 2.1.37 (There exist infinitely many $n$ such that $n!+1$ is divisible by at least two distinct primes)

Take $n=p-1$, where $p>5$ is a prime number (there are infinitely many such integers $n$). By Wilson’s theorem, $p\mid n!+1=(p-1)!+1$, and by exercise 36, $n!+1=(p-1)!+1>1$ is not a power of $p$. Therefore $n!+1$ has two prime factors $p$ and another prime factor, otherwise $n!+1=p^k$ is a power of $p$.

There exist infinitely many $n$ such that $n!+1$ is divisible by at least two distinct primes.