What's the formula to solve summation of logarithms?

By using the fact that $$\log a + \log b = \log ab $$ then

$$ \sum^n \log i = \log (n!) $$

$$ \sum^n \ln i = \ln (n!) $$

Since $\log(A)+\log(B)=\log(AB)$, then $\sum_{i=1}^n\log(i)=\log(n!)$. I'm not sure if this helps a lot since you have changed a summation of $n$ terms into a product of $n$ factors, but it's something.