Does $p^n$ divide $\binom{p^{n+m-1}}{m}$?

Question

Let $n, m \in \mathbf N$ and $p$ an odd prime number. Then does $p^n$ divide $\binom{p^{n+m-1}}{m}$ ?

It seems true, but I can not find a clue. Can I have any hint?

Answer 1

Hint $$\binom{p^{n+m-1}}{m}=\dfrac{p^{n+m-1}}{m}\binom{p^{n+m-1}-1}{m-1}$$

Answer 2

I post an answer following math110's hint.
To complete the proof, we have to show $\frac{p^{m-1}}{m} \binom{p^{n+m-1}-1}{m-1}$ is an integer.
Let $m = p^r a$ where $a$ does not have a factor $p$.
It suffices to show that $r \leq m-1$.
If then, $$\frac{p^{m-1}}{m} \binom{p^{m-1}-1}{m-1}$$ $$= \frac{p^{m-1-r}}{a} \binom{p^{n+m-1}-1}{m-1}.$$
Since $a$ has no factor of $p$, $a$ must divide $\binom{p^{n+m-1}-1}{m-1}$.

Lemma. If $p$ is an integer $\ge 2$, then $p^r \ge r+1$.

proof. For $r=0$, it is obvious.
Using induction on $r$, for $r \ge 1$, $p^r = p p^{r-1} \ge 2r = r +r \ge r+1$.

By the lemma, $m = p^r a \ge (r+1)a \ge r+1$.

$\mathbf{QED}$