Closed form of $\sum_{n = 1}^{\infty} \frac{n^{n - k}}{e^{n} \cdot n!}$ When seeing this question I noticed that
$$
\sum_{n = 1}^{\infty} \frac{n^{n - 2}}{e^{n} \cdot n!}
= \frac{1}{2}.
$$
I don't know how to show this, I tried finding a power series that matches that but no avail.
Hints are very much appreciated.
But this can be generalised:
Define
$$
S_{k}(x)
:= \sum_{n = 1}^{\infty} \frac{n^{n - k}}{x^{n} \cdot n!}.
$$
WolframAlpha shows e.g. that $S_1(e) = 1$ and
$$
S_0(x) = - \frac{W(-x^{-1})}{1 + W(-x^{-1})},
$$
where $W$ denotes the Lambert W-function.
Is there any closed form for this sum or a special $k$ or $x$ beyond those results?
Somebody attempted to answer this using the Lagrange inversion theorem. I didn't work out completely but looked quite promising.
 A: For each $k \geq 1$ we define $T_k$ by
$$T_k(z) := \sum_{n\geq 1} \frac{n^{n-k}}{n!}z^n.$$
We make several observations.


*

*$T_1(z) = -W(-z)$ from the Taylor series of the Lambert $W$-function, which is
$$
W(z)
= \sum_{n \geq 1} (-n)^{n-1} \frac{z^n}{n!}.
$$
In particular, we get
$$T_1(we^{-w}) = w $$
for all $w \leq 1$. As a special case, we get $T_1(e^{-1}) = 1$.

*$T_k'(z) = T_{k-1}(z)/z$. In particular,
$$ \frac{\mathrm{d}}{\mathrm{d}w} T_k(w e^{-w}) = \frac{1-w}{w}T_{k-1}(w e^{-w}). $$
From this, we can compute $T_2$ as follows:
\begin{align*}
T_2(we^{-w})
= \int \frac{1-w}{w}T_{1}(w e^{-w}) \, \mathrm{d}w
= w - \frac{w^2}{2}.
\end{align*}
In particular,
$$ \sum_{n\geq 1} \frac{n^{n-2}}{n! e^n} = T_2(e^{-1}) = \frac{1}{2}. $$
A similar argument shows that each $T_k(we^{-w})$ is a degree $k$ polynomial over $\mathbb{Q}$ without constant term, which can be computed recursively in $k$. For instance, we have
$$
\begin{array}{|c|ccccccc|}
\hline
k & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
T_k(e^{-1}) & 1 & \dfrac{1}{2} & \dfrac{5}{12} & \dfrac{7}{18} & \dfrac{1631}{4320} & \dfrac{96547}{259200} & \dfrac{40291823}{108864000} \\
\hline
\end{array}
$$

Addendum - closed form of $P_n$.
As above, let $P_n$ be the polynomial such that $T_n(z) = P_n(T_1(z))$. Then we claim that the following generating function identity holds.

$$ \sum_{n=1}^{\infty} P_n(z)t^n = - \sum_{n=1}^{\infty} \prod_{i=1}^{n} \frac{tz}{t-i}. \tag{*} $$

Once we prove this, we can find the coefficient of $z^k$ in $P_n(z)$ as
$$ [z^k] P_n(z)
= - [t^n] \frac{t^k}{(t-1)(t-2)\cdots(t-k)}
= \frac{1}{k!} \sum_{j=1}^{k} (-1)^{k-j} \binom{k}{j} \frac{1}{j^{n-k}}. $$
Consequently, we get
$$ P_n(z) = \sum_{k=1}^{n} \Bigg( \sum_{j=1}^{k} (-1)^{k-j} \binom{k}{j} \frac{1}{j^{n-k}} \Bigg) \frac{z^k}{k!}. $$
We conclude with the proof of $\text{(*)}$.
Proof of $\text{(*)}$. Using partial fraction decomposition, we find that
\begin{align*}
- \frac{t^k}{(t-1)\cdots(t-k)}
&= -1 - \sum_{j=1}^{k} (-1)^{k-j} \frac{j^k}{(j-1)!(k-j)!} \frac{1}{t-j} \\
&= -1 + \sum_{j=1}^{k} \frac{(-1)^{k-j}}{j!(k-j)!} j^k \sum_{n=0}^{\infty} \frac{t^n}{j^n}.
\end{align*}
Multiplying $z^k$ and summing over $k = 1, 2, \cdots$, we get
\begin{align*}
- \sum_{k=1}^{\infty} \frac{t^k z^k}{(t-1)\cdots(t-k)}
&= -\frac{z}{1-z} + \sum_{k=1}^{\infty} z^k \sum_{j=1}^{k} \frac{(-1)^{k-j}}{j!(k-j)!} j^k \sum_{n=0}^{\infty} \frac{t^n}{j^n} \\
&= -\frac{z}{1-z} + \sum_{n=0}^{\infty} \sum_{j=1}^{\infty} \frac{t^n}{j! j^n} \sum_{k=j}^{\infty} z^k \frac{(-1)^{k-j}}{(k-j)!} j^k \\ \\
&= -\frac{z}{1-z} + \sum_{n=0}^{\infty} \sum_{j=1}^{\infty} \frac{t^n}{j!} j^{j-n} (ze^{-z})^j \\
&= -\frac{z}{1-z} + \sum_{n=0}^{\infty} T_n(ze^{-z}) t^n.
\end{align*}
Now the desired equality follows by noting that
$$ T_0(ze^{-z}) = \frac{z}{1-z} \qquad \text{and} \qquad T_n(ze^{-z}) = P_n(z) $$
for $n \geq 1$.
