$\newcommand{\blfloor}{\bigl\lfloor}$
$\newcommand{\brfloor}{\bigl\rfloor}$

I have found a way to solve this using Granville's hint.

We have $n!=(n!)_pp^{\blfloor\dfrac{n}{p}\brfloor}\big({\blfloor\dfrac{n}{p}\brfloor}\big)!$, hence
$$\binom{n}{m}=\dfrac{(n!)_p}{(m!)_p(r!)_p}\dfrac{p^{\blfloor\dfrac{n}{p}\brfloor}}{p^{\blfloor\dfrac{m}{p}\brfloor}p^{\blfloor\dfrac{r}{p}\brfloor}}.\dfrac{\big({\blfloor\dfrac{n}{p}\brfloor}\big)!}{\big({\blfloor\dfrac{m}{p}\brfloor}\big)!\big({\blfloor\dfrac{r}{p}\brfloor}\big)!}.$$
Repeating the above process, we get
$$\dfrac{\big({\blfloor\dfrac{n}{p}\brfloor}\big)!}{\big({\blfloor\dfrac{m}{p}\brfloor}\big)!\big({\blfloor\dfrac{r}{p}\brfloor}\big)!}=\dfrac{\big({\blfloor\dfrac{n}{p}\brfloor}!\big)_p}{\big({\blfloor\dfrac{m}{p}\brfloor}!\big)_p\big({\blfloor\dfrac{r}{p}\brfloor!}\big)_p}.\dfrac{p^{\blfloor\dfrac{n}{p^2}\brfloor}}{p^{\blfloor\dfrac{m}{p^2}\brfloor}p^{\blfloor\dfrac{r}{p^2}\brfloor}}\dfrac{\big({\blfloor\dfrac{n}{p^2}\brfloor}\big)!}{\big({\blfloor\dfrac{m}{p^2}\brfloor}\big)!\big({\blfloor\dfrac{r}{p^2}\brfloor}\big)!}.$$

Therefore, by induction, we can claim that
$$\binom{n}{m}=\frac{(n!)_p}{(m!)_p(r!)_p}\prod_{i=1}^s\dfrac{\big({\blfloor\dfrac{n}{p^i}\brfloor}!\big)_p}{\big({\blfloor\dfrac{m}{p^i}\brfloor}!\big)_p\big({\blfloor\dfrac{r}{p^i}\brfloor}!\big)_p}.\prod_{i=1}^s\dfrac{p^{\blfloor\dfrac{n}{p^i}\brfloor}}{p^{\blfloor\dfrac{m}{p^i}\brfloor}p^{\blfloor\dfrac{r}{p^i}\brfloor}}.$$

From Granville's paper, we know that $\blfloor \dfrac{n}{p^i}\brfloor-\blfloor \dfrac{m}{p^i}\brfloor-\blfloor \dfrac{r}{p^i}\brfloor=\epsilon_i$, where $\epsilon_i$ is $1$ if there is a carry entering position $i$, and $0$ otherwise.
Thus, the third factor in the above product is

$$p^{\sum_{i=1}\left(\blfloor \dfrac{n}{p^i}\brfloor-\blfloor\dfrac{m}{p^i}\brfloor-\blfloor\dfrac{r}{p^i}\brfloor\right)}=p^{\sum_{i=0}^s\epsilon_i};$$

by Kummer's theorem this is the highest power of $p$ dividing $\binom{n}{m}$.

Hence,
$$\dfrac{1}{p^k}\binom{n}{m}=\frac{(n!)_p}{(m!)_p(r!)_p}\prod_{i=1}^s\dfrac{\big({\blfloor\dfrac{n}{p^i}\brfloor}!\big)_p}{\big({\blfloor\dfrac{m}{p^i}\brfloor!}\big)_p\big({\blfloor\dfrac{r}{p^i}\brfloor!}\big)_p}.$$

We know that $(-1)^{\blfloor \dfrac{n}{p} \brfloor}(n!)_p\equiv n_0!\ (\text{mod}\ p)$, so
$$\dfrac{\big({\blfloor\dfrac{n}{p^i}\brfloor}!\big)_p}{\big({\blfloor\dfrac{m}{p^i}\brfloor}!\big)_p\big({\blfloor\dfrac{r}{p^i}\brfloor!}\big)_p}\equiv (-1)^{\blfloor \dfrac{n}{p^i} \brfloor-\blfloor \dfrac{m}{p^i} \brfloor-\blfloor \dfrac{r}{p^i} \brfloor}\dfrac{n_i!}{m_i!r_i!}=(-1)^{\epsilon_{i}}\dfrac{n_i!}{m_i!r_i!}.$$

Applying this result also the first factor gives
$$\dfrac{1}{p^k}\binom{n}{m}\equiv \prod_{i=0}^s (-1)^{\epsilon_{i}}\dfrac{n_i!}{m_i!r_i!}=(-1)^k\dfrac{n_0!}{m_0!r_0!}\dfrac{n_1!}{m_1!r_1!}...\dfrac{n_s!}{m_s!r_s!}.$$

Proof is completed.