Prove that $1+\sum_{n\geq1}\sum^n_{k=1}\frac{n!}{k!}{n-1\choose k-1}x^k\frac{u^n}{n!}=\exp\frac{xu}{1-u}$ Let $k,n\in\mathbb{N}_{>0}$.  How do I get started to prove that"
$$1+\left(\sum_{n\geq1}\sum^n_{k=1}\frac{n!}{k!}{n-1\choose k-1}x^k \frac{u^n}{n!}\right) = \exp\frac{xu}{1-u}$$
Hints and help greatly appreciated!
 A: Use this theorem from Enumerative Combinatorics Volume 2 by R. Stanley,

A: $\newcommand{\+}{^{\dagger}}
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$\ds{1 + \sum_{n \geq 1}u^{n}
     \bracks{\color{#00f}{\sum_{k = 1}^{n}{1 \over k!}{n - 1 \choose k - 1}x^{k}}}
     =\exp\pars{xu \over 1 - u}:\ {\large ?}}$

\begin{align}&\color{#00f}{\sum_{k = 1}^{n}{1 \over k!}{n - 1 \choose k - 1}x^{k}}
=\sum_{k = 0}^{\infty}{1 \over k!}\bracks{%
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n - 1} \over z^{k}}
\,{\dd z \over 2\pi\ic}}x^{k}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}\pars{1 + z}^{n - 1}
\bracks{\sum_{k = 0}^{\infty}{\pars{x/z}^{k} \over k!}}\,{\dd z \over 2\pi\ic}
=\color{#00f}{\oint_{\verts{z}\ =\ 1}\pars{1 + z}^{n - 1}\exp\pars{x \over z}
\,{\dd z \over 2\pi\ic}}
\end{align}

\begin{align}&1 + \sum_{n \geq 1}u^{n}
\bracks{\color{#00f}{\sum_{k = 1}^{n}{1 \over k!}{n - 1 \choose k - 1}x^{k}}}
=1 + \oint_{\verts{z}\ =\ 1}{\exp\pars{x/z} \over 1 + z}
\sum_{n \geq 1}\bracks{u\pars{1 + z}}^{n}\,{\dd z \over 2\pi\ic}
\\[3mm]&=1 + \oint_{\verts{z}\ =\ 1}{\exp\pars{x/z} \over 1 + z}
\bracks{{1 \over 1 - u\pars{1 + z}} - 1}\,{\dd z \over 2\pi\ic}
=1 + \oint_{\verts{z}\ =\ 1}\exp\pars{x/z}\,{u \over 1 - u - uz}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=1  + \color{#f00}{{u \over 1 -u}\,{\rm res}\pars{%
\exp\pars{x \over z}\,{1 \over 1 - uz/\pars{1 - u}}}_{z\ = 0}}
\qquad\qquad\qquad\qquad\qquad\qquad\quad\pars{1}
\end{align}

\begin{align}&\color{#f00}{{u \over 1 -u}\,
{\rm res}\pars{%
\exp\pars{x \over z}\,{1 \over 1 - uz/\pars{1 - u}}}_{z\ = 0}}
\\[3mm]&={u \over 1 - u}\,{\rm res}\,\pars{%
\sum_{\ell = 0}{x^{\ell} \over \ell!}\,{1 \over z^{\ell}}
\sum_{\ell' = 0}^{\infty}\pars{u \over 1 - u}^{\ell'}z^{\ell'}}_{z\ =\ 0}
=\sum_{\ell = 1}{1 \over \ell!}\,\pars{{u \over 1 - u}\,x}^{\ell}
\\[3mm]&=\exp\pars{xu \over 1 - u} - 1
\end{align}
  Replace this result in $\pars{1}$ and we arrive to the result.

In order to arrive to the OP result we assumed that
$\ds{\verts{1 - u \over u} > 1}$ such that the only pole is at $\ds{z = 0}$. In addition some intermediate sums require $\ds{\verts{u} < \half}$. I didn't study the compatibility of both conditions. If the OP say something else about $\ds{u}$ I can recheck my answer.
