If $p$ is prime then $p^{n-1}\mid \binom{p^n}{p}$. I am stuck with this problem:  

If $p$ is prime then $p^{n-1}\mid \binom{p^n}{p}$.

The thing is that I don't know much properties of binomial coefficients and I'd accepts hints. 

By Kummer, the $p$-adic valuation of $\binom{p^n}{p}$ is equal to the number of "carry-overs" when adding $p^n-p$ and $p$ in base $p$. The base $p$ expression of $p^n-p$ is $$0\cdot p^n +(p-1)\cdot p^{n-1}+\cdots+(p-1)\cdot p^2+(p-1)\cdot p+0\cdot p^0$$
so adding $p^n-p$ and $p$ in base $p$ yields $n-1$ carry overs (one from the $p^1$ term, then this gives a carry over from the $p^2$ term, and so on until we get a carry over from the $p^{n-1}$ term).
