Prove (or derive) the de Polignac formula for the prime decomposition of $n!$

I can't seem to find any papers published dedicated to show that the de Polignac formula has a rigorous derivation. From Wikipedia's entry for the formula:

Let $n \geq 1$ be an integer. The prime decomposition of $n!$ is given by $$n! = \prod_{\text{prime } p \leq n}\;p^{s_p(n)}$$ where $$s_p(n) = \sum_{j=1}^\infty \left\lfloor \frac{n}{p^j} \right\rfloor $$

Could someone please let me know the derivation or provide a link to a paper which sheds light upon the derivation of the aforementioned formula?

Thank you in advance.

Solution 1:

In the product $1\cdot2\cdot3\cdots n$, there are exactly $m:=\lfloor\dfrac np\rfloor$ factors that are multiples of a given prime $p$, namely $p,2p,3p,\cdots mp$.

Discarding all other factors and dividing by $p^m$, what remains is $m!$, and you can iterate.

Example:

With $p=3$, $$14!=1\cdot2\cdot\color{green}3\cdot4\cdot5\cdot\color{green}6\cdot7\cdot8\cdot\color{green}9\cdot10\cdot11\cdot\color{green}{12}\cdot13\cdot14$$ then $$4!=1\cdot2\cdot\color{green}3\cdot4,$$

so that the multiplicity of $3$ is $5$.

Solution 2:

I find it easier to think of the (exponent of) the power of $p$ dividing $n!$ as $\frac{n- \sigma_{p}(n)}{p-1}$, where $\sigma_{p}(n)$ is the sum ( in $\mathbb{Z}$ ) of the digits of the base $p$ expansion of $n$. This is the same as stated in the question, but expressed differently.

It isn't really necessary, but it is possible to see this using a little group theory. If we write $n = a_{0} + a_{1}p + \ldots + a_{s}p^{s}$ where $0 \leq a_{i} \leq p-1$ for each $i$, it is easy to check that the power of $p$ dividing $n!$ is $a_{1}d_{1} + \ldots + a_{s}d_{s}$ where $p^{d_{i}}$ is the power of $p$ dividing $p^{i}!$.

Using a wreath product construction from group theory, it is possible to see that the power of $p$ dividing $p^{i}!$ is $\frac{p^{i}-1}{p-1}$ for each $i$. Alternatively, it is possible to prove directly by induction that $d_{i+1} = 1 + pd_{i}$ for each $i$, and it is clear that $d_{1} = 1$, so the value of $d_{i}$ is as claimed for each $i$.

Hence the (exponent in) power of $p$ diving $n!$ is $\sum_{i=1}^{n} \frac{a_{i}p^{i}- a_{i}}{p-1}$, which is $\frac{n- \sigma_{p}(n)}{p-1}$, as claimed above.