How to compute the following sum? How to compute the following sum?
$$\sum_{k=1}^{\infty} \frac{k^{k-1} \cdot e^{-k}}{k!}$$
It is likely to be equal $1$ (there is an argumentation that goes back to random graphs). Moreover, i think that $e^{-1}$ can be replaced with $ce^{-c}$ for many c.
 A: To compute this series, let us look at 
$$
\sum_{k=1}^{\infty} \frac{k^{k-1}}{k!} (- z e^z)^k = \sum_{k=1}^{\infty} \frac{(-1)^k k^{k-1}}{k!} z^k e^{kz}
$$
as a formal power series. This series can be rewritten in the form
$$
\sum_{m=1}^{\infty} \alpha_m z^m.
$$ 
where 
\begin{align*}
\alpha_m &= \sum_{k=1}^m \frac{(-1)^k \, k^{k-1}}{k!} (\text{coef. by }z^{m-k} \text{ in Taylor expansion of }e^{kz}) \\&= \sum_{k=1}^m \frac{(-1)^k \, k^{k-1}}{k!} \frac{k^{m-k}}{(m-k)!} = \frac{1}{m!} \sum_{k=1}^m {m \choose k} (-1)^k k^{m-1}.
\end{align*}
It is not too difficult to show that $\alpha_m = 0$ for $m \geq 2$. One way to show is to use the inclusion/exclusion principle to compute the number of $m-1$ long words that use at least one of each $m$ letters. This number is clearly zero, but using the inclusion/exclusion principle over the subsets ($A_i$, the subset of $m-1$ long words that do not use the $i$'th letter) you'll get the right hand side sum above. 
Consequently, we have that 
$$
\sum_{k=1}^{\infty} \frac{k^{k-1}}{k!} (-z e^z)^k = 0 - z + 0z^2 + 0z^3+ \ldots,
$$
as long as $-z e^z$ is in the interval of convergence for the power series $\sum_k \frac{k^{k-1}}{k!} x^k$. Plugging in $z=-1$ gives the formula for the series that you want. 
