Proving $p\nmid \binom{p^rm}{p^r}$ where $p\nmid m$
Let's try another proof using polynomials that I think is cool :
Consider the polynomial $$ (1+X)^p - (1+X^p) $$ over the field $\mathbb{Z}_p$. We know that $a^p \equiv a\pmod{p}$ for all $a\in \mathbb{Z}_p$, so this is a polynomial of degree $p-1$, that has $p$ roots. This is only possible if it is identically zero. Hence, $$ (1+X)^p \equiv (1+X^p)\pmod{p} $$ By induction, we see that $$ (1+X)^{p^r} \equiv (1+X^{p^r})\pmod{p} $$ Thus $$ (1+X)^{p^rm} \equiv (1+X^{p^r})^m\pmod{p} $$ Hence the corresponding coefficients of both these polynomials must be equal $\pmod{p}$. Consider the coefficient of $X^{p^r}$ on both sides to conclude that $$ {p^r m \choose p^r } \equiv m\pmod{p} $$ which proves what you want.
Here is a different approach.
For a prime $p$, the Corollary below says that the number of factors of $p$ that divide $\binom{n}{k}$ is $$ \frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}\tag{1} $$ where $\sigma_p(n)$ is the sum of the digits in the base-$p$ represention of $n$.
For any $m$, $\sigma_p\left(mp^r\right)=\sigma_p\left(m\right)$. Therefore, $$ \frac{\sigma_p\left(p^r\right)+\sigma_p\left((m-1)p^r\right)-\sigma_p\left(mp^r\right)}{p-1} =\frac{\sigma_p(1)+\sigma_p(m-1)-\sigma_p(m)}{p-1}\tag{2} $$ Thus, if $p\not\mid m$, the low order digit of $m$ in base-$p$ is not $0$ and so the base-$p$ representation of $m-1$ is the same as the base-$p$ representation of $m$ with the lowest order digit reduced by $1$. Thus the quantity on the right side of $(2)$ is $0$.
QED
Lemma (Legendre's Formula): The number of factors of a prime $p$ that divide $n!$ is $$ \frac{n-\sigma_p(n)}{p-1}\tag{3} $$ Proof: Note that the number of multiples of $p^j$ not greater than $n$ is $\left\lfloor\frac n{p^j}\right\rfloor$. Thus, to compute the number of factors of $p$ that divides $n!$ we count the number of multiples of $p$ not greater than $n$, then add the number of multiples of $p^2$ not greater than $n$ (we add them once since they have already been counted once with the multiples of $p$), then add the number of multiples of $p^3$ not greater than $n$ (we add them once since they have already been counted once with the multiples of $p$ and once with the multiples of $p^2$), and so forth. That is, the number of factors of $p$ in $n!$ is $$ \sum_{j=1}^k\left\lfloor\frac n{p^j}\right\rfloor\tag{4} $$ where $k$ is the number of digits in the base-$p$ representation of $n$.
Now suppose that $$ n=\sum_{i=0}^kd_ip^i\tag{5} $$ Then $$ \begin{align} \sum_{j=1}^k\left\lfloor\frac n{p^j}\right\rfloor &=\sum_{j=1}^k\sum_{i=j}^kd_ip^{i-j}\\ &=\sum_{i=1}^k\sum_{j=1}^id_ip^{i-j}\\ &=\sum_{i=1}^kd_i\frac{p^i-1}{p-1}\\ &=\sum_{i=0}^kd_i\frac{p^i-1}{p-1}\\ &=\frac{n-\sigma_p(n)}{p-1}\tag{6} \end{align} $$ QED
Corollary (Kummer's Theorem): The number of factors of $p$ that divide $\binom{n}{k}$ is $$ \frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}\tag{7} $$ Proof: Note that $$ \binom{n}{k}=\frac{n!}{k!(n-k)!}\tag{8} $$ and apply the Lemma to each term in the numerator and denominator of $(8)$.
QED
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.