Proving formula for $p$-adic ordinal of $n!$ 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
 A: You can first note a different formula: 
$$\operatorname{ord}_p (n!) = \sum_{i=1}^{\infty} \lfloor n/p^i \rfloor.$$
Now if you write $n= \sum_{i=0}^{\ell} n_i p^i$ you get that the above sum actually only goes up to $\ell$ (then the summands are $0$) and  the $j$-th term is 
$$\sum_{i=j}^{\ell} n_i p^{i-j}.$$
Thus you have expressed what you want as a double sum 
$$\sum_{j=1}^{\ell}  \sum_{i=j}^{\ell} n_i p^{i-j}.$$
Now, rearrange summation, or look with which powers of $p$ each $n_i$ appears. 
You find $n_i(1 + \dots + p^{i-1})=  n_i(p^i -1)/(p-1)$. 
So in total you have 
$$
\sum_{i=1}^{\ell} n_i(p^i -1)/(p-1) = (p-1)^{-1} \sum_{i=0}^{\ell} (n_ip^i -n_i ) = (p-1)^{-1}( n -  \sum_{i=0}^{\ell} n_i) 
$$
as you want. 
This leaves to assert the starting representation, but this follows by noting that 
$\lfloor n/p^1 \rfloor$ is the number of $m\le n$ divisible by $p$,
$\lfloor n/p^2 \rfloor$ is the number of $m\le n$ divisible by $p^2$, and so on. 
A: 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.
