What's the formula to solve summation of logarithms?

I'm studying summation. Everything I know so far is that:

$\sum_{i=1}^n\ k = \frac{n(n+1)}{2}\ $

$\sum_{i=1}^{n}\ k^2 = \frac{n(n+1)(2n+1)}{6}\ $

$\sum_{i=1}^{n}\ k^3 = \frac{n^2(n+1)^2}{4}\ $

Unfortunately, I can't find neither on my book nor on the internet what the result of:

$\sum_{i=1}^n\log i$.

$\sum_{i=1}^n\ln i$.

is.

Can you help me out?

Answer

By using the fact that

$$\log a + \log b = \log ab$$ then

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

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