I've seen related questions, but none quite like this one. For a rational number $r$ let $[r]$ be the largest integer less than or equal to $r$, e.g., $[\frac{1}{2}] = 0, [2] = 2$, and $[3\frac{1}{3}] = 3$. Prove $v_p n! = [n/p] + [n/p^2]+[n/p^3]+\cdots$. My professor encourages us to use this new function: Let $\#_{p,k}(n)$ be the number of positive integers less than or equal to $n$ that are divisible by $p^k$.
He then also recommended I make this new $d$ function, where $d_{pk}(j) = \{1$ if $p^k|j, 0$ otherwise$\}$.
I then started to question my work here:
$$v_p(n!) = \sum_{m=1}^n v_p(m) = \sum_{m=1}^n\sum_{i=1}^md_{p,v_p(i)}(i)$$ $$=n(\sum_{i=1}^1d_{p,v_p(i)}(i))+(n-1)(\sum_{i=2}^2d_{p,v_p(i)}(i))+\cdots+ (1)(\sum_{i=n}^nd_{p,v_p(i)}(i))$$ $$=\sum_{k=1}^{v_p(n)}\#_{p,k}(n) = \sum_{k=1}^{\infty}\#_{p,k}(n) = [\frac{n}{p}]+[\frac{n}{p^2}]+\cdots$$
Besides the fact that this is quite abstract and confusing, I am unsure about this double summation I have. If this is incorrect, how should I proceed?