Saudi Arabia Mathematical Competitions

Since $2011$ is a prime and $2011^{2011}$ divides $n!$, it follows that in $n!$ we have at least $2011$ multiples of $2011$. These are $$2011,2\cdot 2011,3\cdot 2011,\dots ,2011\cdot 2011.$$

Therefore $2011!\cdot 2011^{2011}$ divides $n!$, hence $2011^{2012}$ divides $n!$.