Who's Wilson?

Consider $n!$ , where $n > 1$ and $(n + 1)$ is a prime integer. By Wilson's Theorem $((n + 1) - 1)! \equiv n! \equiv -1 \bmod (n + 1)$ which shows that $(n + 1)\nmid n!$ . Thus, for the given problem, if $n = 11$ , $11 \nmid 10!$ , which gives $\boxed{\text{No}}$ .

By the Fundamental Theorem of Arithmetic, every integer can be expressed uniquely by a combination of prime numbers. Since $10!$ can only be expressed by the prime numbers $2,3,5,7$ , then $11$ which is itself a prime number is not included there. Since no multiple of 11 is present in $10!$ , it does not divide $10!$ .

11 is a prime number