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?

$$\log(1-x)=-\sum_{n=1}^\infty \frac{1}{n}x^n$$
so
$$\log(x)=-\log\frac{1}{x}=-\log(1-\frac{x-1}{x})=\sum_{n=1}^\infty \frac{1}{n}(\frac{x-1}{x})^n$$
And
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$
so
$$x=\frac{1}{1-\frac{x-1}{x}}=\sum_{n=0}^\infty (\frac{x-1}{x})^n$$
and thus
$$x\log(x)=(\sum_{n=1}^\infty \frac{1}{n}(\frac{x-1}{x})^n)\cdot \sum_{m=0}^\infty (\frac{x-1}{x})^m = \sum_{n=1}^\infty a_n(\frac{x-1}{x})^n$$
where
$$a_n=\sum_{k=0}^{n-1} \frac{1}{n-k}=1+\frac{1}{2}+\frac{1}{3}+....\frac{1}{n}=\frac{S_n}{n!}$$
So the result is
$$x\log(x)=\sum_{n=1}^\infty \frac{S_n}{n!}(\frac{x-1}{x})^n$$
