How many times does p appears in this sequence I'm trying to understand the proof for this theorem, page 233. One of the steps is to prove that every term in this expression:
$$
\frac{p^\alpha}{j} \binom{p^\alpha}{j-1} k^{j-1}p^{(j-1)\beta}, \;\;\; (j>1)
$$
is divisible by $p^\alpha$. As $p^\alpha$ appears in the numerator and $\binom{p^\alpha}{j-1}$ and $ k^{j-1}p^{(j-1)\beta}$ are integers, $j$ is the only factor which can "take" some part of $p^\alpha$. Now we have to prove that $p^{(j-1)\beta}$ divides $j$. To do that, it says that the number of times the factor $p$ can appear in $j\in\{2, \ldots, p^\alpha\}$, is:
$$
\frac{j}{p}+\frac{j}{p^2}+\frac{j}{p^3}+\cdots=\frac{j}{p-1}
$$
But the factor $p$ appears at least this many times in $p^{(j-1)\beta}$, since $\beta\geq1$.
Those are the two statements I cannot fully understand. How has he counted that? I do not know where that formula comes from. Any help?

This is a bound of the Legendre's formula i.e.,
$$\nu _p(n!)=\sum _{k=1}^{\infty}\left\lfloor \frac{n}{p^k}\right \rfloor =\frac{n-s_p(n)}{p-1},$$
where $s_p(n)$ is the sum of the digits of $n$ in base $p.$
The bound is by dropping the ceiling function i.e., $\lfloor x\rfloor \leq x$ and geometric series.
