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?
 A: 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
$\begin{align} x\ln x &= \dfrac{1}{1-u}   \ln \left(\dfrac{1}{1-u}\right) = -\dfrac{1}{1-u}   \ln (1-u)\\
&=\left(1+u+u^2 + \cdots\right)\left(u+\frac{u^2}{2}+\frac{u^3}{3} + \cdots\right)\\
&=u+(\frac{1}{2}+1)u^2+(\frac{1}{3}+\frac{1}{2}+1)u^3+(\frac{1}{4}+\frac{1}{3} + \frac{1}{2}+1)u^4+\cdots\\
&=u+H_2u^2+H_3u^3+H_4u^4+\cdots\\
&=u+\frac{S(2,2)}{2!}u^2+\frac{S(3,2)}{3!}u^3+\frac{S(4,2)}{4!}u^4+\cdots
+\frac{S(n,2)}{n!}u^n+\cdots\\
\end{align}$
A: 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 $$\sum_{m = 0}^\infty \frac{S_m}{m!}\left(\frac{x-1}{x}\right)^{m+1}$$
A: 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.
A: Admittedly the same is done in one form or another in previous answers, I just wanted to make it as concise as possible.
Substituting $x=\frac1{1-y}$, what we have to prove is$$(1+y+...+y^n+...)(y+\frac{y^2}2+...+\frac{y^n}n+...)=y+\frac3{2!}y^2+...+\frac{S_n}{n!}y^n+...,$$i. e. $n!(1+\frac12+...+\frac1n)=S_n$.
Now what you denote by $S_n$ is usually denoted $|s(n+1,2)|$ or $\left[\matrix{n+1\\2}\right]$, and the needed equality is explicitly given e. g. in Wikipedia.
Just for fun, here is a quick combinatorial proof. By definition $S_n$ is the number of permutations of $n+1$ consisting of exactly two (disjoint) cycles. To name such a permutation means to name a subset of $\{1,...,n+1\}$ of size, say, $k$, with $0<k<n+1$, together with cyclic ordering of this subset and of its complement.
Now the number of subsets of size $k$ is $\binom{n+1}k$, while the number of cyclic orderings of a set of size $\ell$ is $(\ell-1)!$, so we get $\binom{n+1}k(k-1)!(n-k)!=\frac{(n+1)!}{k(n+1-k)}=n!\left(\frac1k+\frac1{n+1-k}\right)$. We then have to sum over $1\le k\le n$ and divide by 2 (since we have counted each disjoint pair of cycles twice, first at $k$ and then at $n+1-k$). The result is $n!H_n$.
A: $$\log(1-x)=-\sum_{n=1}^\infty \frac{1}{n}x^n$$
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
$$x\log(x)=\sum_{n=1}^\infty \frac{S_n}{n!}(\frac{x-1}{x})^n$$
