For each positive integer $n$, let $n!$ denote the product $1\cdot 2\cdot 3\cdot\,\cdots\,\cdot (n-1)\cdot n$.

What is the remainder when $9!$ is divided by $11$?
I recall that Wilson's theorem says that if $p$ is a prime number, then $(p-1)!\equiv -1\pmod{p}$. This means that the remainder when $(p-1)!$ is divided by $p$ is $p-1$. I can apply this theorem to the case $p=11$, and get that $10!\equiv -1\pmod{11}$. Now, I can divide both sides by 10, and use the fact that the inverse of 10 modulo 11 is 10, since $10\cdot 10=100\equiv 1\pmod{11}$. This gives me that $9!\equiv -10\equiv 1\pmod{11}$. Therefore, the remainder when 9! is divided by 11 is 1. 