James Stirling (1692-1770, Scotland)
![{\displaystyle \left.\ln N!\right.=\ln 1+\ln 2+\ln 3+...+\ln N=\sum _{k=1}^{N}\ln k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ce04a543290ba3e91a5482d38e0898061309f7d)
Because of [Euler-MacLaurin formula]
![{\displaystyle \sum _{k=1}^{N}\ln k=\int _{1}^{N}\ln x\,dx+\sum _{k=1}^{p}{\frac {B_{2k}}{2k(2k-1)}}\left({\frac {1}{n^{2k-1}}}-1\right)+R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/155c1159d5843fd787bc4cf5ff01571d8a25c1cc)
where B1 = −1/2, B2 = 1/6, B3 = 0, B4 = −1/30, B5 = 0, B6 = 1/42, B7 = 0, B8 = −1/30, ... are the Bernoulli numbers, and R is an error term which is normally small for suitable values of p.
![{\displaystyle ~\approx \int _{1}^{N}\ln xdx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80d303178931d236619dedc5c834941dbd8495fd)
![{\displaystyle ~=\left[x\ln x-x\right]_{1}^{N}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ba6f8a16f83420e34962f787fd0668a13baa710)
![{\displaystyle ~=N\ln N-N+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a749b2acbeb696c1f3f1544d34dd67fa0edb676)
Thus, for large N
![{\displaystyle \ln N!\approx N\ln N-N}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1aad6782caeef37ddaf2ef6cff7ddbc9ef4848a)