I need the proving of $x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$ I could find the role of Striling Numbers  in the natural logarithm function as follows

$$x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$$
  Where $S_n$= absolute Striling Numbers of first kind (0,1,3,11,50,274....)

this series numerically checked without any problem, but  I need the proving. Any help?

We have 
$$x\log(x) = x[-\log(1/x)] = x\left[-\log\left(1 - \frac{x-1}{x}\right)\right] = x\sum_{m = 0}^\infty \frac{1}{m+1}\left(\frac{x-1}{x}\right)^{m+1},$$
and $$x = \dfrac{1}{\frac{1}{x}} = \dfrac{1}{1 - \frac{x-1}{x}} = \sum_{m = 0}^\infty \left(\frac{x - 1}{x}\right)^m.$$ Thus
$$x\log(x) = \sum_{m = 0}^\infty \left(\frac{x-1}{x}\right)^m \sum_{m = 0}^\infty \frac{1}{m}\left(\frac{x-1}{x}\right)^m = \sum_{m = 0}^\infty \sum_{k = 0}^m \frac{1}{k+1}\left(\frac{x-1}{x}\right)^{m+1},$$
which is the same as $$\sum_{m = 0}^\infty \frac{S_m}{m!}\left(\frac{x-1}{x}\right)^{m+1}$$
