Highest power of a prime $p$ dividing $N!$ How does one find the highest power of a prime $p$ that divides $N!$ and other related products?
Related question: How many zeros are there at the end of $N!$?

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For a number $n$, define $\Lambda(n)=\log p$ if $n=p^k$ and zero elsewhen. 
Note that $\log n=\sum\limits_{d\mid n}\Lambda(d)$. If $N=n!$, then $$\log N=\sum_{k=1}^n\log k=\sum_{k=1}^n\sum_{d\mid k}\Lambda (d)=\sum_{d=1}^n \Lambda(d)\left\lfloor \frac nd\right\rfloor$$
Since $\Lambda(d)$ is nonzero precisely when $d$ is a power of a prime and in such case it equals $\log p$, the last sum equals $$\sum_{p}\sum_{k\geqslant 1}\log p \left\lfloor\frac n{p^k}\right\rfloor$$ and this gives $$\nu_p(n!)=\sum_{k\geqslant 1}\left\lfloor\frac n{p^k}\right\rfloor$$
If you write $n$ is base $p$, say $n=a_0+a_1p+\cdots+a_kp^k$, the above gives that $$\nu_p(n!)=\frac{s(n)-n}{1-p}$$ where $s(n)=a_0+\cdots+a_k$. 
