Let $n\in\mathbb{N}$. Can you find a simple expression for the function
$f_n(z)=\sum_{k=0}^\infty \frac{k^nz^k}{k!}$.
By simple expression I just mean a finite linear combination of powers of $z$, exponential functions, (or something else?)
For example in the case $n=1$, we have
$f_1(z)=ze^z$.
I am really struggling to see an expression for the general $n\in\mathbb{N}$ case. Any help would be much appreciated.
On
\begin{align} f_{n+1}(z)&=\sum_{k\ge 1}\frac{k^{n+1}z^k}{k!}=z\sum_{k\ge 0}\frac{(k+1)^{n}z^{k}}{k!}\\ &=z\sum_{i=0}^n\sum_{k\ge 0}\binom{n}{i}\frac{k^iz^{k}}{k!} \\ &=z\sum_{i=0}^n\binom{n}{i}f_i(z). \end{align}
On
A similar trick should work always$$(ze^z)'= \sum \frac{k(k-1)z^k}{k!} = \sum \frac{k^2z^k}{k!} - \sum \frac{kz^k}{k!} = \sum \frac{k^2z^k}{k!} - ze^z.$$
On
In terms of Stirling numbers of the second kind for any $n\geq 1$ we have $$ k^n = \sum_{j=1}^{n}{n \brace j}\binom{k}{j}j! \tag{1}$$ and since $$ \sum_{k\geq 0}\frac{z^k}{k!}\binom{k}{j}j! = z^j e^z \tag{2}$$ it follows that $$ \sum_{k\geq 0}\frac{k^n z^j}{k!} = e^z\sum_{j=1}^{n}{n\brace j}z^j\tag{3} $$ so the wanted series is the product between $e^z$ and a Bell polynomial.
Consider $z=e^x$. Then we have:
$$g_n(x)=f_n(e^x)=\sum_{k=0}^\infty\frac{k^ne^{kx}}{k!}\\g_0(x)=e^{e^x}$$
Likewise, one can easily see that this is the $n$th derivative of $g_0(x)$:
$$k^ne^{kx}=\frac{\partial^n}{\partial x^n}e^{kx}$$
Thus,
$$g_n(x)=\frac{\partial^n}{\partial x^n}e^{e^x}$$
And by Faà di Bruno's formula,
$$g_n(x)=\sum_{k=1}^n e^{e^x}\cdot B_{n,k}(\underbrace{e^x,e^x,\dots,e^x}_{n-k+1})$$
Where $B_{n,k}$ is a Bell polynomial. Now just substitute $x=\ln(z)$ to get
$$f_n(z)=\sum_{k=1}^n e^z\cdot B_{n,k}(\underbrace{z,z,\dots,z}_{n-k+1})$$
This is also known as a Touchard polynomial. We have:
$$f_n(z)=e^zT_n(z)=e^z\sum_{k=0}^n\left\lbrace {n \atop k}\right\rbrace z^k=e^z\sum_{k=0}^n\frac1{k!}\sum _{j=0}^k(-1)^{k-j}\binom kjj^nz^k$$
Where $\left\lbrace {n \atop k}\right\rbrace$ is a Stirling number of the second kind.