Proof. For the proof we can allow of numbers in base $p$ with leading zeros. So let \begin{eqnarray*} n_dn_{d-1}\cdots n_0 & := n, \\ m_dm_{d-1}\cdots m_0 & := m, \end{eqnarray*}all in base $p$ We set $r=n-m$ and denote the $p$ adic representation of $r$ with $r_dr_{d-1}\cdots r_0$

We define $c_{-1} =0$ and for each $0 \le j \le d$ \begin{equation} c_j =\begin{cases} 1 & \text{for $m_j +r_j \ge p$} \\ 0 &\text{otherwise.} \end{cases} \end{equation} Finally, we introduce $\delta_p(n)$ as the sum of digits in the $p$ adic of $n$ Then it follows that the power of $p$ dividing $\binom{n}{m}$ is $$\frac{\delta_p(m) +\delta_p(r) -\delta_p(n)}{p-1}.$$ For each $j \ge 0$ we have $$n_j =m_j +r_j +c_{j-1} -p.c_j.$$ Then \begin{eqnarray*} \delta_p(m) +\delta_p(r) -\delta_p(n) &=\sum_{k=0}^d \left(m_k +r_k -n_k\right) \\ &=\sum_{k=0}^d \left((p-1)c_j\right) &+\sum_{k=0}^d \left(c_j -c_{j-1}\right) \\ &=\sum_{k=0}^d (p-1)c_j&+c_d -c_{-1} \end{eqnarray*}This gives us $$\frac{\delta_p(m) +\delta_p(r) -\delta_p(n)}{p-1} =\sum_{k=0}^d c_k,$$ the total number of carries.