What powers of $p$, where $p$ is prime, divide $\binom {p^{\alpha} m}{p^{\alpha}}$?

Question

My question is about a short combinatorial discussion in Herstein's Topics in Algebra, on page 92, preceding one of his three proofs of Sylow's Theorem. I will cite the important part:

If $p^r | m$ but $p^{r+1} \nmid m$, consider $$\binom {p^\alpha m}{p^{\alpha}} = \frac{(p^{\alpha}m)!}{(p^{\alpha})!(p^{\alpha}m-p^{\alpha})!} = \frac {p^{\alpha}m(p^{\alpha}m-1) \dots (p^{\alpha}m-i) \dots (p^{\alpha}m-p^{\alpha} + 1)} {p^{\alpha}(p^{\alpha}-1) \dots (p^{\alpha}-i) \dots (p^{\alpha}-p^{\alpha} + 1)}.$$ The question is, What power of p divides $\binom {p^{\alpha} m}{p^{\alpha}}$? Looking at this number, written out as we have written it out, one can see that except for the term $m$ in the numerator, the powers of $p$ dividing $(p^{\alpha}m-i$$)$ is the same as that dividing $p^{\alpha}-i$, so all powers of $p$ cancel out except the power which divides $m$. Thus $p^r | \binom {p^{\alpha} m}{p^{\alpha}}$ but $p^{r+1} \nmid \binom {p^{\alpha} m}{p^{\alpha}}$.

My understanding is as follows. $\binom {p^\alpha m}{p^{\alpha}} = m *\frac{\binom {p^\alpha m}{p^{\alpha}}}{m}$, consider the unique factorization of both factors. The right will not contain any nonzero powers of $p$, and so the power of $p$ dividing our binomial is the largest power in the unique factorization of $m$. But there are two gaps in my understanding: why is the second factor neccesarily an integer, and why does the second factor not contain powers of $p$? Herstein explains this second question, but I don't get his argument.

Answer 1

Observe that $\displaystyle\binom{p^\alpha m}{p^\alpha} = m \displaystyle\binom{p^\alpha m - 1}{p^\alpha - 1}$, so the second factor is an integer.

Now the claim is that the factors of $p$ in $\displaystyle\binom{p^\alpha m - 1}{p^\alpha - 1}$ all cancel out perfectly from the numerator and denominator. This will follow if we show that, for all $i$ with $1 \leq i \leq p^\alpha - 1$, a power of $p$ (say $p^k$) divides $p^\alpha m - i$ if and only if it divides $p^\alpha - i$.

Proposition: Let $1 \leq i \leq p^\alpha - 1$. Let $k$ be a positive integer. (i) If $k > \alpha$, then $p^k \nmid p^\alpha - i$ and $p^k \nmid p^\alpha m - i$. (ii) If $1 \leq k \leq \alpha$, then $p^k \mid p^\alpha - i$ if and only if $p^k \mid p^\alpha m - i$.

Proof: (i) Since $k > \alpha$, $p^k > p^\alpha > p^\alpha - i$ so $p^k \nmid p^\alpha - i$. And $p^k \nmid p^\alpha m - i$, otherwise if say $p^\alpha m - i = p^k l$, then $i = p^\alpha m - p^kl = p^\alpha (m - p^{k - \alpha}l)$ (note $k - \alpha > 0$) so $p^\alpha \mid i$, a contradiction since $i \leq p^\alpha - 1 < p^\alpha$.

(ii) Suppose $p^\alpha - i = p^kl$ for some $l$, so $i = p^\alpha - p^kl = p^k(p^{\alpha - k} - l)$ (note $\alpha - k \geq 0$) so $p^k \mid i$. Combining this with $p^k \mid p^\alpha m$, we get $p^k \mid p^\alpha m - i$. Conversely suppose $p^\alpha m - i = p^k l$ for some $l$. So $i = p^\alpha m - p^k l= p^k(p^{\alpha - k}m - l)$ (note $\alpha - k \geq 0$), so $p^k \mid i$. Combine this with $p^k \mid p^\alpha$, we get $p^k \mid p^\alpha - i$. $\square$

Answer 2

Kummer's Theorem, which is proven in this answer, says that the number of factors of $p$ in $\binom{n}{k}$ is the number of carries when adding $k$ to $n-k$ in base-$p$.

If $p^r\mid m$ and $p^{r+1}\nmid m$, then adding $1$ to $m-1$ will have exactly $r$ base-$p$ carries. Likewise, adding $p^\alpha$ to $(m-1)p^\alpha$ will have exactly $r$ base-$p$ carries. Therefore, there will be $r$ factors of $p$ in $\binom{mp^\alpha}{p^\alpha}$.