How do I find the closed form sum formula of
$$\displaystyle\sum_{k=1}^{\log_2(n)}\log_2 k,$$
I was trying to do $\log_2 1 + \log_2 2 + \cdots + \log_2(\log_2 n)$ and using property of $\log$: $\log a + \log b = \log(ab)$ to get $$\log_2(1\cdot 2\cdot 3 \cdots (\log_2 n)) = \log_2((\log_2 n)!)$$ but I don't think it's right... Is there a better formula?
Since sum indices only go over integer values, you only take the sum up to $\lfloor\log_2 n\rfloor$. That being the case, I think $\log_2(\lfloor\log_2 n\rfloor!)$ is about the best exact answer you are going to get. You could use stirling's approximation to estimate this value as $$\lfloor\log_2 n\rfloor\log_2\left(\frac{\lfloor\log_2 n\rfloor}{e}\right)+\frac{1}{2}\log_2\left(2\pi\lfloor\log_2 n\rfloor\right)$$ but I don't know how helpful that really is, and it won't be quite as accurate for $\ln(n!)$ as it is for $n!$.