TITLE: To prove that $\sum\limits_{n=1}^{\infty}\frac{n^{n-1}}{n!e^n}=1$
QUESTION [7 upvotes]: $\sum\limits_{n=1}^{\infty}\frac{n^{n-1}}{n!e^n}=1?$
I have noticed that $\lim\limits_{n\to\infty}{\frac{1}{\sqrt[n]\frac{n^{n-1}}{n!}}}=1/e$, so the series $\sum\limits_{n=1}^{\infty}\frac{n^{n-1}}{n!}x^n$converges in $[-\frac1e,\frac1e]$, then I suppose to find the sum-function of $\sum\limits_{n=1}^{\infty}\frac{n^{n-1}}{n!}x^n$. However, this is surely difficult for me. Could anyone help me out? Sincerely thanks for your help.

REPLY [5 votes]: The function 
$$W(z) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} n^{n-1} z^n}{n!}$$
is called the Lambert $W$ function (see also Wikipedia), so you are trying to compute $-W(-e^{-1})$. As Wikipedia says, $W(-e^{-1}) =-1$ so your claimed identity is right. More generally,
$$z = W(z) e^{W(z)}$$
so 
$$-e^{-1} = W(-e^{-1}) e^{W(-e^{-1})},$$
which strongly suggests the correct result $W(-e^{-1}) =-1$.
This is already mostly said in par's answer. I write further because your language suggests that you think that a power series with ratio $r$ necessarily converges on the closed interval $[-r,r]$. In fact, you don't know whether or not there is convergence at the end points. There is in this case, but it isn't obvious.