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?

Suppose we seek to show that
$$x\log x = \sum_{n=1}^\infty \frac{1}{n!} \left[n+1\atop 2\right]
\left(\frac{x-1}{x}\right)^n.$$

Recall the species equation for decompositions into disjoint cycles:
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}(\mathcal{Z}))$$
which gives the generating function
$$G(z, u) = \exp\left(u\log\frac{1}{1-z}\right)$$
so that
$$\left[n+1\atop 2\right]
= \frac{1}{2} (n+1)! [z^{n+1}] \left(\log\frac{1}{1-z}\right)^2.$$

Substitute this into the sum to get
$$\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n!} 
(n+1)! \left(\frac{x-1}{x}\right)^n  
[z^{n+1}] \left(\log\frac{1}{1-z}\right)^2
\\ = \frac{1}{2} \sum_{n=1}^\infty
(n+1) \left(\frac{x-1}{x}\right)^n  
[z^{n+1}] \left(\log\frac{1}{1-z}\right)^2.$$
Now $\left(\log\frac{1}{1-z}\right)^2$ starts at  $z^2$ so that we may
include the value for $n=0$ in the sum (zero contribution) to get
$$\frac{1}{2} \sum_{n=0}^\infty
(n+1) \left(\frac{x-1}{x}\right)^n  
[z^{n+1}] \left(\log\frac{1}{1-z}\right)^2.$$
What we have here is an annihilated coefficient extractor that 
evaluates to
$$\frac{1}{2} 
\left. \frac{d}{dz} \left(\log\frac{1}{1-z}\right)^2
\right|_{z=(x-1)/x}.$$
This is
$$\left. \frac{1}{1-z} \log\frac{1}{1-z}
\right|_{z=(x-1)/x}$$
which finally yields
$$\frac{1}{1-(x-1)/x} \log\frac{1}{1-(x-1)/x}
= \frac{x}{x-(x-1)} \log\frac{x}{x-(x-1)}
\\ = x \log x.$$
There is another annihilated coefficient extractor at this 
MSE link I and another one at this 
MSE link II and at this MSE link III.
