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!

  • $\begingroup$ Are $n$ and $m$ the same thing? $\endgroup$ Apr 10, 2016 at 1:26
  • $\begingroup$ Indeed, thanks for pointing that out! $\endgroup$
    – C. Falcon
    Apr 10, 2016 at 1:27

1 Answer 1


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 :

$$ v_{p}(m!) = \sum_{i=1}^{m} \frac{m}{p^{i}} \leqslant m \cdot\left(\frac{1}{1-\frac{1}{p}} -1\right) = \frac{m}{p-1} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.