Let $\epsilon_p(n)=\lfloor {n/p}\rfloor+ \lfloor {n/p^2}\rfloor+\cdots$ i.e the largest poswer of $p$ (prime) that divides $n!$ where $n$ is an integer.
Let $(\alpha_0\ldots\alpha_m)$ be a epresentation of $n$ with respect to the base $p$.
How can we show that
$ n!/p^{\epsilon_p(n)}\equiv (-1)^{\epsilon_p(n)} \alpha_0!\ldots\alpha_m!\ (\textrm{mod } p) ? $
Any help is really appreciated.