TITLE: Proving formula for $p$-adic ordinal of $n!$
QUESTION [7 upvotes]: Let $n \geq 1$ and $n = n_0 + ... n_{\ell}p^{\ell}$ be the p-adic expansion of $n$. Define $\alpha_p(n) = n_0 + ... n_\ell$.  Then $\mathrm{ord}_p(n!)= \frac{n - \alpha_p(n)}{p-1}$.  I'm trying to prove this by induction and that doesn't seem to work without insanity.  Any ideas?
P-adic Notes by Andrew Baker

REPLY [5 votes]: Since $\mathrm{ord}_p(n!)$ is only affected by the numbers divisible by $p$ we see that:
$$\begin{align}
\mathrm{ord}_p(n!) &= \mathrm{ord}_p\left((p\cdot 1)(p\cdot 2)\dots \left(p\cdot\left\lfloor\frac{n}{p}\right\rfloor\right)\right)
\\&= \left\lfloor\frac{n}{p}\right\rfloor + \mathrm{ord}_p\left(\left\lfloor\frac{n}{p}\right\rfloor!\right)
\end{align}$$
So the induction from $n$ to $n+1$ is hard, but the induction from $\left\lfloor\frac{n}{p}\right\rfloor$ to $n$ is relatively easy.
Basically, the induction is on $\ell$, not $n$.
Note that $\left\lfloor\frac{n}{p}\right\rfloor = \frac{n-n_0}p$, and, by the induction hypothesis:
$$\mathrm{ord}_p\left(\left\lfloor\frac{n}{p}\right\rfloor!\right) = \frac{1}{p-1}\left(\frac{n-n_0}{p} - \sum_{i=1}^\ell n_i\right) = \frac{n-n_0}{p(p-1)} - \frac{\sum_{i=1}^\ell n_i}{p-1}$$
Now adding $\frac{n-n_0}{p} = \left\lfloor\frac{n}{p}\right\rfloor$ to both sides and you are done.