# Majoration of the $p$-adic valuation of a factorial.

Let $p$ be a prime number.

In order to prove a result on $p$-adic interpolation of iterates, I need to show the following:

Lemma. Let $m$ be an integer, one has: $$v_p(m!)\leqslant\frac{m}{p-1}.$$

At that moment I only know that: $$v_p(m!)=\sum_{i=1}^mv_p(i).$$

If possible I would prefer to avoid using Legendre's formula. Thank you in advance for your help!

• Are $n$ and $m$ the same thing? – carmichael561 Apr 10 '16 at 1:26
• Indeed, thanks for pointing that out! – C. Falcon Apr 10 '16 at 1:27

From : $$v_p(m!)=\sum_{i=1}^mv_p(i).$$
We can obtain, for a big enough $n$ : $$v_{p}(m!) = \left| \left\{k \mid pk \leqslant m \right\} \right| + \left| \left\{k \mid p^{2}k \leqslant m \right\} \right| + ... + \left| \left\{k \mid p^nk \leqslant m \right\} \right|$$
Indeed, if $i$ is such that $v_p(i) = m$, i will be in the $m$ first sets.
Thus we get : $$v_{p}(m!) = \left| \left\{k \mid k \leqslant \frac{m}{p} \right\} \right| + \left| \left\{k \mid k \leqslant \frac{m}{p^{2}} \right\} \right| + ... +\left| \left\{k \mid k \leqslant \frac{m}{p^{n}} \right\} \right|$$ And enventually :
$$v_{p}(m!) = \sum_{i=1}^{m} \frac{m}{p^{i}} \leqslant m \cdot\left(\frac{1}{1-\frac{1}{p}} -1\right) = \frac{m}{p-1}$$