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 will need the fact that $$n!H_n = n!\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} \right) = S(n,2)\\ \text{ where } S(n,2) \text{ is the Stirlings number of the first kind}$$
let me make a change of variable $u= \dfrac{x-1}{x}, x = \dfrac{1}{1-u}.$ then
$\begin{align} x\ln x &= \dfrac{1}{1-u}   \ln \left(\dfrac{1}{1-u}\right) = -\dfrac{1}{1-u}   \ln (1-u)\\
&=\left(1+u+u^2 + \cdots\right)\left(u+\frac{u^2}{2}+\frac{u^3}{3} + \cdots\right)\\
&=u+(\frac{1}{2}+1)u^2+(\frac{1}{3}+\frac{1}{2}+1)u^3+(\frac{1}{4}+\frac{1}{3} + \frac{1}{2}+1)u^4+\cdots\\
&=u+H_2u^2+H_3u^3+H_4u^4+\cdots\\
&=u+\frac{S(2,2)}{2!}u^2+\frac{S(3,2)}{3!}u^3+\frac{S(4,2)}{4!}u^4+\cdots
+\frac{S(n,2)}{n!}u^n+\cdots\\
\end{align}$
