Originally Posted by
prasum how did you get this
To expand on Also sprach Zarathustra's post in general if $\displaystyle p$ is prime then the $\displaystyle m$ for which $\displaystyle p^m\| n!$ (this means that $\displaystyle p^m\mid n!$ but $\displaystyle p^{m+1}\nmid n!$) is $\displaystyle \displaystyle \sum_{k=1}^{\infty}\left\lfoor \frac{n}{p^k}\right\rfloor$. To understand why this is try to understand why
$\displaystyle \displaystyle m=\left(\#\text{ of }k\leqslant n!\text{ such that }p\mid k\right)+\left(\#\text{ of }k\leqslant n!\text{ such that }p^2\mid k\right)+\left(\#\text{ of }k\leqslant n!\text{ such that }p^3\mid k\right)+\cdots$
and how this sum describes that number.