De Polignac's Formula

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

Let $$p$$ be a prime number.

Then $$p^\mu$$ is a divisor of $$n!$$, and $$p^{\mu + 1}$$ is not, where:
 * $$\mu = \sum_{k > 0} \left \lfloor{\frac {n}{p^k}}\right \rfloor$$

where $$\left \lfloor{\cdot}\right \rfloor$$ denotes the Floor Function.

Proof
Note that although the summation given in the statement of the theorem is given as an infinite sum, in fact it terminates after a finite number of terms (because when $$p^k > n$$ we have $$0 < n/p^k < 1$$).

From Number of Multiples Less Than a Given Number, we have that $$\left \lfloor{\frac {n}{p^k}}\right \rfloor$$ is the number of integers $$m$$ such that $$0 < m \le n$$ which are multiples of $$p^k$$.

We look more closely at $$n!$$:
 * $$n! = 1 \times 2 \times \ldots \times \left({n-1}\right) \times n$$

We see that any integer $$m$$ such that $$0 < m \le n$$ which is divisible by $$p^j$$ and not $$p^{j+1}$$ must be counted exactly $$j$$ times.

That is: once in $$\left \lfloor{\frac {n}{p}}\right \rfloor$$, once in $$\left \lfloor{\frac {n}{p^2}}\right \rfloor$$, $$\ldots$$, once in $$\left \lfloor{\frac {n}{p^j}}\right \rfloor$$.

And that is all the occurrences of $$p$$ as a factor of $$n!$$.

Thus:
 * $$\mu = \left \lfloor{\frac {n}{p}}\right \rfloor + \left \lfloor{\frac {n}{p^2}}\right \rfloor + \cdots + \left \lfloor{\frac {n}{p^j}}\right \rfloor$$

Hence the result.