$v(n!)＝\sum_{i=1}^{\infty} [\dfrac{n}{p^i}]v(p)$

Question

Let $R$ be a discrete valuation ring that is complete with respect to its maximal ideal $M$,and let $v$ be the valuation on $R$. And $p$ be a prime. And we assume $0＜v(p)＜∞$.

Why equality $v(n!)＝\sum_{i=1}^{\infty} [\dfrac{n}{p^i}]v(p)$ holds?

I know this is true if $v$ is p-adic valuation, this case is elementary.

Thank you in advance.

Answer 1

$n! = \prod_{q^k} q^{\lfloor n/q^k \rfloor }$

$v(p)>0$ gives that $v(q)=0$ for $q\ne p$.