Nice equality involving summation, factorials, and $e$ I came across this nice equality that I don't know how to solve:
$$\sum_{j=1}^{\infty} \frac{e^{-j}j^{j-1}}{j!}=1.$$
I'm guessing the first step is to write out the Taylor series, which gives a double sum of:
$$\sum_{j=1}^{\infty} \frac{\left(\sum_{n=1}^{\infty}\frac{(-j)^n}{n!}\right)j^{j-1}}{j!}$$
and maybe this can be seen to be the product of two other functions in Taylor form, which I can't see. Also, I know this can be solved by elementary methods, so a "simple" solution would be nice. Help is appreciated!
 A: Hint. For $|x|< 1/e$, the power series 
$$T(x)=\sum_{j=1}^{\infty}\frac{j^{j-1}}{j!}\,x^j,$$
which is the exponential generating function for rooted labeled trees,
satisfies the identity
$$T(x)=xe^{T(x)}.$$
Now you need to show that $T(1/e)=1$.
A: By Lagrange inversion theorem, the principal branch of the Lambert $W$ function, defined through $W(x)e^{W(x)}=x$, has the following expansion near the origin:
$$ W(x) = \sum_{n\geq 1}\frac{n^{n-1}(-1)^{n-1}}{n!}x^n\tag{1} $$
In particular,
$$ \sum_{n\geq 1}\frac{n^{n-1}}{n! e^{n}} = -W\left(-\frac{1}{e}\right) = \color{red}{1}\tag{2} $$
since the function $x\cdot e^x$ equals $-\frac{1}{e}$ at $x=-1$, from which $W\left(-\frac{1}{e}\right)=-1$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{j\ =\ 1}^{\infty}{\expo{-j}j^{\, j - 1} \over j!} & =
\sum_{j\ =\ 1}^{\infty}{\expo{-j} \over j!}\,\,\,
\overbrace{\pars{j - 1}!
\oint_{\verts{z}\ =\ 1}{\expo{\, jz} \over z^{\, j}}\,{\dd z \over 2\pi\ic}}
^{\ds{=\ j^{\, j - 1}}}\,\,\, =\,\,\,
\oint_{\verts{z}\ =\ 1}\,\,\,\sum_{j\ =\ 1}^{\infty}
{\pars{\expo{z - 1}/z}^{\, j} \over j}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
-\oint_{\verts{z}\ =\ 1}\,\,\,
\ln\pars{1 - {\expo{z - 1} \over z}}\,{\dd z \over 2\pi\ic} =
-1 -
\oint_{\verts{z}\ =\ 1}\,\,\,
\ln\pars{z - \expo{z - 1}}\,{\dd z \over 2\pi\ic}
\end{align}
The equation $\ds{z - \expo{z - 1} = 0}$ has infinite solutions$^{\large\S}$.
One of them is obviously $\ds{z = 1}$ and the remaining ones have
$\ds{\verts{z} > 1}$.
$$\bbox[#ffd,10px,border:0.5px groove navy]
{\begin{array}{l}
{\large\S}:~\mbox{They are evaluated by}\ \texttt{Mathematica}\ \mbox{as}\
\texttt{ProductLog[k,-1/e]}\ \mbox{with}\,\,\, k \in \mathbb{Z}\
\end{array}}
$$
It reduces the complex integration to:
\begin{align}
\sum_{j\ =\ 1}^{\infty}{\expo{-j}j^{\, j - 1} \over j!} & =
-1\ -\,\,\, \overbrace{\oint_{\verts{z}\ =\ 1}\,\,\,
\ln\pars{z - 1}\,{\dd z \over 2\pi\ic}}^{\ds{=\ -2}} = \bbx{\ds{1}}
\end{align}
