jmc

Let the base 11 expansion of $10!$ be $a_n{a}_{n-1}\cdots a_1{a}_0$. This implies that $10!=11^n{a}_n+11^{n-1}{a}_{n-1}+\cdots 11a_1+a_0$. Note that $10!$ is not divisible by $11$, because $11$ is prime. As a result, if $a_0=0$, then the right-hand side of that equation would be divisible by $11$, which is a contradiction. Therefore, $a_0\ne 0$, and $10!$ ends in $\boxed{0}$ zeroes when written in base 11.