elementary number theory - highest power of prime $p$ dividing $binom{m+n}{n}$

How to prove the theorem stated here.

Theorem. (Kummer, 1854) Let $p$ be a prime. The highest power of $p$ that divides the binomial coefficient $\binom{m+n}{n}$ is equal to the number of "carries" when adding $m$ and $n$ in base $p$.

So far, I know if $m+n$ can be expanded in base power as
$$m+n= a_0 + a_1 p + \dots +a_k p^k$$
and $m$ have to coefficients $\{ b_0 , b_1 , \dots b_i\}$ and $n$ can be expanded with coefficients $\{c_0, c_1 ,\dots , c_j\}$ in base $p$ then the highest power of prime that divides $\binom{m+n}{n}$ can be expressed as
$$e = \frac{(b_0 + b_1 + \dots b_i )+ (c_0 + c_1 + \dots c_j )-(a_0 + a_1 + \dots a_k )}{p-1}$$
It follows from here page number $4$. But how does it relate to the number of carries? I am not being able to connect. Perhaps I am not understanding something very fundamental about addition.

Answer

If $b_{0} + c_{0} < p$, then $a_{0} = b_{0} + c_{0}$, there are no carries, and the term

$$
b_{0} + c_{0} - a_{0} = 0
$$
does not contribute to your $e$.

If $b_{0} + c_{0} \ge p$, then $a_{0} = b_{0} + c_{0} - p$, and this time $b_{0} + c_{0} - a_{0}$ gives a contribution of $p$ to the numerator of $e$. Plus, there is a contribution of $1$ to $a_{1}$, so the net contribution to the numerator of $e$ is $p -1$, and that to $e$ is $1$. Repeat.

As mentioned by Jyrki Lahtonen in his comment (which appeared while I was typesetting this answer), you may have propagating carries, but this is the basic argument.