Highest power of a prime $p$ dividing $N!$ How does one find the highest power of a prime $p$ that divides $N!$ and other related products?
Related question: How many zeros are there at the end of $N!$?

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Here's a different approach I found while thinking in terms of relating the sum of digits of consecutive numbers written in base $p$. Part of the appeal of this approach is you might have at one point learned that there is a connection but don't remember what, this gives a quick way to reconstruct it.
Let's consider $s_p(n)$, the sum of digits of $n$ in base $p$. How does the sum of digits change if we add $1$ to it? It usually just increments the last digit by 1, so most of the time,
$$s_p(n+1) = s_p(n)+1$$
But that's not always true. If the last digit was $p-1$ we'd end up carrying, so we'd lose $p-1$ in the sum of digits, but still gain $1$. So the formula in that case would be,
$$s_p(n+1) = s_p(n) + 1 - (p-1)$$
But what if after carrying, we end up carrying again? Then we'd keep losing another $p-1$ term. Every time we carry, we leave a $0$ behind as a digit in that place, so the total number of times we lose $p-1$ is exactly the number of $0$s that $n+1$ ends in, $v_p(n+1)$.
$$s_p(n+1) = s_p(n) + 1 - (p-1)v_p(n+1)$$
Now let's rearrange to make the telescoping series and sum over,
$$\sum_{n=0}^{k-1} s_p(n+1) - s_p(n) = \sum_{n=0}^{k-1}1 - (p-1)v_p(n+1)$$
$$s_p(k) = k - (p-1) \sum_{n=0}^{k-1}v_p(n+1)$$
Notice that since $v_p$ is completely additive so long as $p$ is a prime number,
$$s_p(k) = k - (p-1) v_p\left(\prod_{n=0}^{k-1}(n+1)\right)$$
$$s_p(k) = k - (p-1) v_p(k!)$$
This rearranges to the well-known Legendre's formula,
$$v_p(k!) = \frac{k-s_p(k)}{p-1}$$
