Factorial Divisible by Prime Power

Theorem
Let $$n \in \Z: n \ge 1$$.

Let $$p$$ be a prime number.

Let $$n$$ be expressed in base $p$ notation:
 * $$n = \sum_{j=0}^m r_j p^j$$

where $$0 \le r_j < p$$.

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

Let $$p^\mu$$ be the largest power of $$p$$ which divides $$n!$$, that is:
 * $$p^\mu \backslash n$$
 * $$p^{\mu+1} \nmid n$$

Then:
 * $$\mu = \frac {n - s_p \left({n}\right)} {p-1}$$

where $$s_p \left({n}\right)$$ is the digit sum of $$n$$.

Proof
If $$p > n$$, then $$s_p \left({n}\right) = n$$ and we have that:
 * $$\frac {n - s_p \left({n}\right)} {p-1} = 0$$

which ties in with the fact that $$\left \lfloor{\frac {n}{p}}\right \rfloor = 0$$.

Hence the result holds for $$p > n$$.

So, let $$p \le n$$.

From Multiplicity of Prime Factor in Factorial, we have that:

$$ $$

where $$p^s < n < p^{s+1}$$.

From Quotient and Remainder to Number Base, we have that:
 * $$\left \lfloor{\frac {n}{p^k}}\right \rfloor = \left[{r_m r_{m-1} \ldots r_{k+1} r_k}\right]_p = \sum_{j=0}^{m-k} r_j p^j$$

where $$0 \le r_j < p$$.

Hence:

$$ $$ $$ $$ $$ $$ $$ $$ $$ $$

Hence the result.

History
This result is due to Adrien-Marie Legendre, who published it in Essai sur la Théorie des Nombres (2nd edition) in 1808.