Stirling transform of $(k-1)!$ While reading about combinatorial mathematics, I found this article about the Stirling transform which caught my attention.
So, if I wanted to find the Stirling transform of, for instance, $(k-1)!$, I'd have to solve this sum (using the explicit formula for the Stirling number of the second kind): $$\sum_{k=1}^{n}\left(\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^n(k-1)!\right)$$
It looks complicated. Is it even possible to find the Stirling transform directly from this sum? I don't know how to start. Mathematica gives the answer $(-1)^n\operatorname{Li}_{1-n}(2)$. I'd really like to know how to arrive at that result.
Any ideas or hints will be appreciated.
 A: Note: The double sum expression stated in OPs question is already  a perfectly valid representation of the Stirling transform of the sequence $(a_n)_{n\geq 1}=\left((n-1)!\right)_{n\geq 1}$ with respect to its definition.
So, here we are looking  for a different representation of the Stirling transform of $(a_n)_{n\geq 1}$ which could be regarded as more convenient or simpler according to our needs. Regrettably, a  considerable simplification, e.g. reducing a sum don't presumably exist. But we establish the connection of OPs Stirling transform with the Polylogarithm $\operatorname{Li}_{1-n}(2)$ provided by Mathematica.

Let's denote the Stirling numbers of the second kind with $\begin{Bmatrix}n\\k\end{Bmatrix}$. If $(a_n)_{n\geq 1}$ is a sequence of numbers then the sequence $(b_n)_{n\geq 1}$ with
  \begin{align*}
b_n=\sum_{k=1}^n\begin{Bmatrix}n\\k\end{Bmatrix}a_k\tag{1}
\end{align*}
  is called the Stirling transform of $(a_n)_{n\geq 1}$.

$$$$

Stirling transform of $(a_n)_{n\geq 1}=\big((n-1)!\big)_{n\geq 1}$
According to formula (1) we obtain using an explicit expression for $\begin{Bmatrix}n\\k\end{Bmatrix}$
  \begin{align*}
b_n&=\sum_{k=1}^{n}\frac{1}{k!}\sum_{j=1}^k(-1)^{k-j}\binom{k}{j}j^na_k\\
&=\sum_{k=1}^{n}\frac{1}{k!}\sum_{j=1}^k(-1)^{k-j}\binom{k}{j}j^n(k-1)!\\
&=\sum_{k=1}^n\sum_{j=1}^{k}(-1)^{k-j}\binom{k-1}{j-1}j^{n-1}\tag{2}
\end{align*}

$$$$

Generating functions: We introduce the generating functions $f,g$ with
  \begin{align*}
f(x)&=\sum_{n=1}^{\infty}a_n\frac{x^n}{n!}
=\sum_{n=1}^{\infty}(n-1)!\frac{x^n}{n!}=\sum_{n=1}^{\infty}\frac{x^n}{n}\\
&=-\ln(1-x)\\
g(x)&=\sum_{n=1}^{\infty}b_n\frac{x^n}{n!}
\end{align*}
The representation of (2) in terms of generating functions is according to this Wiki page
  \begin{align*}
g(x)=f(e^x-1)
\end{align*}
  We conclude
  \begin{align*}
g(x)&=f(e^x-1)\\
&=-\ln (2-e^x)\\
&=-\ln 2 -\ln\left(1-\frac{e^x}{2}\right)\tag{3}\\
\end{align*}
  Expanding the logarithmic series we obtain
  \begin{align*}
-\ln\left(1-\frac{e^x}{2}\right)&=\sum_{j=1}^{\infty}\frac{1}{2^j}\frac{e^{jx}}{j}\\
&=\sum_{j=1}^{\infty}\frac{1}{2^jj}\sum_{m=0}^{\infty}\frac{(jx)^m}{m!}\\
&=\sum_{m=0}^{\infty}\left(\sum_{j=1}^{\infty}\frac{1}{2^j}j^{m-1}\right)\frac{x^m}{m!}\\
&=\ln(2)+\sum_{m=1}^{\infty}\left(\sum_{j=1}^{\infty}\frac{1}{2^j}j^{m-1}\right)\frac{x^m}{m!}\tag{4}\\
\end{align*}
Combining (3) and (4) we get
\begin{align*}
g(x)=\sum_{m=1}^{\infty}\left(\sum_{j=1}^{\infty}\frac{1}{2^j}j^{m-1}\right)\frac{x^m}{m!}
\end{align*}
We use the coefficient of operator $[x^m]$ to denote the coefficient of $x^{m}$ in $g(x)$ and obtain
\begin{align*}
b_n&=\frac{1}{n!}[x^n]g(x)=\sum_{j=1}^{\infty}\frac{1}{2^j}j^{n-1}\qquad\qquad n\geq 1\tag{5}
\end{align*}

$$$$

Polylogarithms: According to the definition of the Polylogarithm
\begin{align*}
  \operatorname{Li}_s(x):=\sum_{j=1}^{\infty}\frac{x^j}{j^s}\qquad\qquad |x|< 1, s\in\mathbb{C}
  \end{align*}
we observe the RHS of (5) can be written letting $x=\frac{1}{2}$ and $s=1-n$
\begin{align*}
  b_n=\sum_{j=1}^{\infty}\frac{1}{2^j}j^{n-1}=\operatorname{Li}_{1-n}\left(\frac{1}{2}\right)
  \end{align*}
Since the following identity is valid for $n\geq 1$
\begin{align*}
  \operatorname{Li}_{-n}(x)+(-1)^n\operatorname{Li}_{-n}\left(\frac{1}{x}\right)=0
  \end{align*}
we conclude
\begin{align*}
  b_n=\operatorname{Li}_{1-n}\left(\frac{1}{2}\right)=(-1)^n\operatorname{Li}_{1-n}(2)\qquad\qquad n\geq 1
  \end{align*}
which corresponds with the result of Mathematica.

