Exercise 2.1.53 (There are infinitely many $n$ such that $n!-1$ is divisible by at least two distinct primes)

Proof. The solution is similar to Exercise 37.

Take $n=p-2$, where $p>5$ is a prime number (there are infinitely many such integers $n$). By Exercise 52, $p\mid n!-1=(p-2)!+1$, and $n!-1=(p-2)!+1>1$ 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. □