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!

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 $W\left(-\frac{1}{e}\right)=-1$.
