__Factorizing the factorial__

Let `n` be a natural number; let `p` be a prime;

`m` = the exponent of `p` in the prime factorization of `n`! ;

`s` = the sum of the (`p`-ary) digits of `n`’s representation in base `p`.

__Theorem__ `m` = ( `n` - `s` ) / ( `p` - 1 )

Because of the usual recursive definition of `n`!, an inductive proof over `n` seems indicated. Since for `n` = 0, `m` = 0 ∧ `s` = 0, the theorem holds for `n` = 0.

Consider now the transition from `n` to `n` + 1; let `k` be the exponent of `p` in the prime factorization of `n` + 1.

Because ( `n` + 1 )! = ( `n` + 1 ) ∙ `n`! and `p` is prime

(0) Δ `m` = `k` .

Because ( `n` + 1 ) - `n` = 1

(1) Δ `n` = 1 .

Because the `p`-ary representation of ( `n` + 1 ) ends on exactly `k` zeros

(2) Δ `s` = 1 - `k`∙( `p` - 1 )

Combining (1) and (2) yields

(3) Δ (( `n` - `s` ) / ( `p` - 1 )) = `k` .

Combining (0) and (3) yields the induction step, and thus the proof is completed.