Problem - 4202
Let $n$ be an integer greater than $1$. If none of $1!$, $2!$, $\cdots$, $n!$ has the same remainder when being divided by $n$, show that $n$ is a prime.
First, $n=2, 3$ are both prime. When $n=4$, we have $2!\equiv 3!\pmod{4}$. By # 4196, we know for any composite $n > 4$, we have $(n-1)!\equiv 0\pmod{n}$. This means $(n-1)!\equiv n!\equiv 0\pmod{n}$.
Therefore we conclude the claim holds.