Showing that $ord_p((p^n)!) = 1 + p + \dots + p^{n-1}$ I need to prove that $ord_p((p^n)!) = 1 + p + \dots + p^{n-1}$ ($ord_p$ is the $p-$adic valuation), which is essentially a nasty combinatorics problem. I want to count how many integers from $1$ to $p^n$ have exactly one factor of $p$, how many have exactly two factors of $p$, up to how many have exactly $n$ factors of $p$. Then I take the corresponding sum to get my answer.
I was thinking about using induction to show that total number of $p$'s which appear in the factorizations of the integers from $p^{n-1}$ to $p^n$ is $p^{n-1}$, which will give my desired result. My problem is finding an organized way of listing and counting this large amount of numbers.

hint: $ord_p(n!)=\sum\limits_{i \in \mathbb Z^+} \lfloor n/p^i\rfloor$ is true for all positive integer $n$. 
It is an easy and straight forward result.
