By the ratio test the series $$ \sum_{n\ge0}\frac{n^x}{n!} $$ is convergent, but I know no method to evaluate it.
Since it's a convergent series then my question here is:
Is there a closed form for $\displaystyle\sum_{n\geq 0}\frac{n^x}{n!}$ with $x$ a positive real number?
I don’t think, that there is a closed form. The question should be changed to: "What interesting features does this series have (e.g. what integral representations are known) ?"
A note:
Be $\enspace\displaystyle f_{N,x}(z):= \sum\limits_{n=0}^N {\binom{N}{n}}n^x z^n\enspace $ with $\,x\geq 0\,$ , $\,z\in\mathbb{C}\,$ and $\,N\in\mathbb{N}_0\,$ .
It follows $\enspace\displaystyle \sum\limits_{n=0}^\infty \frac{n^x}{n!} = \lim\limits_{N\to\infty} f_{N,x}(\frac{1}{N}) \,$ .
For $\enspace x\in\mathbb{N}\enspace$ we have $\enspace\displaystyle f_{N,x}(z) = Nz(z+1)^{N-x}\sum\limits_{k=0}^{x-1}z^k \sum\limits_{v=0}^k {\binom{x-N}{k-v}}{\binom{N-1}{v}}(v+1)^{x-1}$
and we see that we get troubles generalizing (e.g) this to real $\,x> 0\,$ .
A hint $\,$ for $\,x\in\mathbb{N}\,$ :
$\displaystyle \sum\limits_{n=0}^\infty \frac{z^n n^x}{n!} = \lim\limits_{N\to\infty} f_{N,x}(\frac{z}{N}) =$
$\enspace\displaystyle = \lim\limits_{N\to\infty} (1+\frac{z}{N})^{N-x} z \lim\limits_{N\to\infty} \sum\limits_{k=0}^{x-1}(\frac{z}{N})^k \sum\limits_{v=0}^k {\binom{x-N}{k-v}}{\binom{N-1}{v}}(v+1)^{x-1}$
$\enspace\displaystyle =e^z \sum\limits_{k=0}^{x} S_{x,k} z^k\enspace$ with the Stirling numbers of the $2^{nd}$ kind $\,\displaystyle S_{x,k}:= \sum\limits_{v=0}^k \frac{(-1)^{k-v} v^x}{(k-v)!v!}\,$ .
The (first) series and the last sum of the equation chain are equal also for $\,x=0\,$ and $\,0^0:=1\,$.
