v_p(n!) from the digits of n in base p

1. Write $n=\sum_{i\ge0}a_ip^i$ with $0\le a_i\le p-1$. Then $\lfloor n/p^k\rfloor=\sum_{i\ge k}a_ip^{i-k}$. By Legendre's formula,
$$v_p(n!)=\sum_{k\ge1}\sum_{i\ge k}a_ip^{i-k}=\sum_{i\ge1}a_i\left(1+p+\dots+p^{i-1}\right)=\sum_{i\ge1}a_i\frac{p^i-1}{p-1}.$$
Since the term $i=0$ contributes $0$, the sum equals $\dfrac{n-s_p(n)}{p-1}$.

2. For $p=2$: $v_2(n!)=n-s_2(n)$. Since $s_2(n)\ge1$, $v_2(n!)\le n-1$, with equality iff $s_2(n)=1$, i.e. $n$ is a power of $2$. So $2^{n-1}\mid n!$ iff $n$ is a power of $2$.

Next, $s_2(2n)=s_2(n)$, so
$$v_2\binom{2n}n=v_2((2n)!)-2v_2(n!)=(2n-s_2(n))-2(n-s_2(n))=s_2(n).$$

Check. $\binom63=20=2^2\cdot5$ and $s_2(3)=2$ ✓. $\binom{8}{4}=70$, $s_2(4)=1$, $v_2(70)=1$ ✓.

Lesson. Legendre's formula is just a statement about the digits of $n$ in base $p$; switching to digits often turns a messy floor sum into a clean closed form.