De Polignac's Formula/Technique

Theorem
Let $n!$ be the factorial of $n$.

Let $p$ be a prime number.

Let $\mu$ be defined as:
 * $\ds \mu = \sum_{k \mathop > 0} \floor {\frac n {p^k} }$

When calculating $\mu$, the easiest way to calculate the next term is simply to divide the previous term by $p$ and discard the remainder:


 * $\floor {\dfrac n {p^{k + 1} } } = \floor {\floor {\dfrac n {p^k} } / p}$

Proof
From Floor of $\dfrac {x + m} n$: Corollary:
 * $\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$

which is valid for all integers $m, n$ such that $n > 0$.

In this instance, $m = 0$ and $n = p$, while $x = \dfrac n {p^k}$.