TITLE: Proving $p\nmid \binom{p^rm}{p^r}$ where $p\nmid m$
QUESTION [7 upvotes]: A question from Advanced Modern Algebra by Joseph J.Rotman.

Let $n=(p^r)m $  such that the prime $p\nmid m$.Prove that $p\nmid \dbinom{n}{p^r}$.HINT: Assume otherwise, cross multiply and apply Euclid's lemma.

I followed the hint and got
$n!=p^r!(n-p^r)!\dbinom{n}{p^r}$
Now using Euclid's lemma, we conclude that since p divides the left side,it must also divide the right side. But we already know from our primary assumption that p divides $\dbinom{n}{p^r}$. But it must also divide $p^r!(n-p^r)!$. I am lost at this point and cannot seem to derive the contradiction as hinted in the hint.
I would greatly appreciate it if you guys could nudge me in the right direction.

REPLY [4 votes]: Using rational numbers and rearranging the way to multiply, it is immediate that:
$$ {p^rm\choose p^r} = \frac {p^rm} {p^r} \cdot \frac {p^rm -1} {p^r -1} \cdot ... \cdot \frac {p^rm -p} {p^r -p} \cdot ... \cdot \frac {p^rm -kp^a} {p^r -kp^a}\cdot ... \cdot\frac {p^rm -p^r+1} {1}$$
$$ = m \cdot \frac {p^rm -1} {p^r -1} \cdot ... \cdot \frac {p^{r-1}m -1} {p^{r-1} -1}\cdot ... \cdot \frac {p^{r-a}m -k} {p^{r-a} -k} \cdot ... \cdot (p^rm -p^r+1) $$
$$ = q_0 \cdot q_1 \cdot ... \cdot q_p \cdot ... \cdot q_{kp^a} \cdot ... \cdot q_{p^r-1}$$
All $q_n$ are free of $p$ in the numerator and denominator when p does not divide m or those k. And therefore $p$ does not divide that combinatorial.