I am quite new to $v_p()$ problems, and would like to know if anyone prove that $v_n(n!)\le n/2$?
Basically, what I mean is that prove that for all positive integers $n$, the amount of factors of $n$ in $n!$ is always going to be at most $n/2$.
For any integers $n \ge 2$ and $m \ge 1$, define $v_n(m) = \max\{k \in \mathbb{N} : n^k \mid m\}$.
Prove that $v_n(n!) \le \dfrac{n}{2}$ for all integers $n \ge 2$.
Note that for any prime $p$, we have $v_p(n!) = \displaystyle\sum_{k = 1}^{\infty}\left\lfloor\dfrac{n}{p^k}\right\rfloor \le \sum_{k = 1}^{\infty}\dfrac{n}{p^k} = \dfrac{n}{p-1}$.
If $n$ is not a power of $2$, then there exists a prime $p \ge 3$ dividing $n$. Since $p \mid n$, we have $v_n(n!) \le v_p(n!) \le \dfrac{n}{p-1} \le \dfrac{n}{2}$.
If $n = 2^m$ for some integer $m \ge 2$, then $v_n(n!) \le \dfrac{v_2(n!)}{m} \le \dfrac{n}{m} \le \dfrac{n}{2}$.
If $n = 2$, we have $v_2(2!) = v_2(2) = 1 = \dfrac{2}{2}$.
Therefore, $v_n(n!) \le \dfrac{n}{2}$ for all integers $n \ge 2$.