Highes power of a prime that divides $n!$ [duplicate]

Question

Hello I am trying to prove that given a prime $p$ and a natural $n$ then the highest power of $p$ that divides $n!$ is:

$V(p)=$ $\sum_{i=1}^{\infty}$ $[n/p^i]$

being []: $ℝ→ℤ$ the function that given any number $x$, returns the largest interger $≤x$.

So I figured I could prove that $p^{V(p)}$ divides $n!$ then prove that there is no greater power $m$ such that $p^m$ divides $n!$. The thing is... I cant seem to prove that $p^{V(p)}$ divides $n!$ , tried by induction, tried to divide it and doing it directly, but in every try there's always a bump I cannot seem to get past.

Appreciate any help.

Answer 1

Hint: Consider the number of factors of $n!$ (when we write $n!=n(n-1)...1)$ of the form $ap$ where $p$ does not divide $a$, then the number of the form $ap^2$ and so on.