After I already got great help with this question I now have the follow-up-question of whether there is a closed form for $$f_n(x) = \sum \limits_{k=1}^{n} \frac{x^k k^k}{k!}$$ or $ \ \ \ \ \ g_n(x) = \sum \limits_{k=0}^{n-1} \frac{x^k (n-k)^k}{k!} \ \ \ \ \ $ respectively for which (negative) $x$ the function $g_n(x)$ eventually becomes negative.
I tried $$g_n(x) = \sum \limits_{k=0}^{n-1} \frac{x^k}{k!} \sum \limits_{j=0}^{k} {{k}\choose{j}} n^{k-j} (-k)^j = \sum \limits_{k=0}^{n-1} \sum \limits_{j=0}^{k} \frac{(-kx)^j}{j!} \frac{(nx)^{k-j}}{(k-j)!} $$
but that led nowhere as well as splitting up $f_n$ as
$$ f_n(x)=\sum \limits_{s=0}^{ \left \lfloor n/2 \right \rfloor} \frac{x^{2s}}{2s!} \left ( (n-2s)^{2s} - \frac{(n-1-2s)^{2s+1} x}{2s+1}\right ). $$
Edit:
Thanks for pointing at the lambert W function, I now came up with $$ f_\infty(x) = - \sum \limits_{n=1}^{ \infty} \sum \limits_{k=n}^{ \infty} \frac{(-k)^{k-1}}{k!} (-x)^k, $$ which is great because x is negative.
By Lagrange inversion formula $$ f_\infty(x) = \sum_{n\geq 1}\frac{n^n x^n}{n!} = -\frac{W(-x)}{1+W(-x)} $$ holds for any $x<e^{-1}$, where $W$ is Lambert function, the inverse function of $xe^x$.
Over such range we simply have $\text{sign}\,f_\infty(x) = \text{sign}(x)$, since $f_\infty$ is increasing and $f_\infty(0)=0$.
The previous formula also tells us that $f_\infty(x)$, despite being given by a not-so-fast-convergent series, can be numerically computed in a pretty effective way through Netwton's method.
As a reference, you may consider the formula $(14)$ here and the Wikipedia page about $W$.
What is the reason leading to the study of the partial sums of such series?