I'm looking for a series of functions like that:$$f_k^{(n)}(x)\mid_{x=0}=n^kn!$$ Here are the first and second functions: $$f_0(x)=\frac 1{1-x}$$$$f_1(x)=\frac x{(1-x)^2}$$ That's because $\frac 1{1-x}^{(n)}\mid_{x=0}=n!\text{ and }\frac x{(1-x)^2}^{(n)}\mid_{x=0}=nn!$
You maybe have notice that those functions are generating function, the first for $\mathbb N$ and the second for $\mathbb N^2$, it's easy to see that $f_k$ is the generating function for $\mathbb N^{k+1}$.
So what I'm actually asking is what is the generating function of:$$\{0^k,1^k,2^k,3^k,\dots\}$$
Since the powers can be expressed as a linear combinations of the Falling Factorials through the Stirling N. of 2nd kind as $$ x^{\,n} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left\{ \matrix{ n \cr k \cr} \right\}\,x^{\,\underline {\,k\,} } } $$ then $$ \eqalign{ & \sum\limits_{0\, \le \,\,k} {k^{\,m} \;z^{\,k} } \quad \left| {\;{\rm 0} \le {\rm integer }m} \right.\quad = \cr & = \sum\limits_{0\, \le \,\,k} {\sum\limits_{0\, \le \,\,j\,\left( { \le \,m} \right)} {\left\{ \matrix{ m \cr j \cr} \right\}k^{\,\underline {\,j\,} } \;z^{\,k} } } = \sum\limits_{0\, \le \,\,j} {\sum\limits_{0\, \le \,\,k} {\left\{ \matrix{ m \cr j \cr} \right\}j!\left( \matrix{ k \cr j \cr} \right)\;z^{\,k} } } = \cr & = \sum\limits_{0\, \le \,j\,\left( { \le \,m} \right)} {\left\{ \matrix{ m \cr j \cr} \right\}} {{j!z^{\,j} } \over {\left( {1 - z} \right)^{\,j + 1} }} = \cr & = \left[ {0 = m} \right] + {z \over {\left( {1 - z} \right)^{m + 1} }}\sum\limits_{0\, \le \,j\,\left( { \le \,m} \right)} {\left\langle \matrix{ m \cr j \cr} \right\rangle z^{\,j} } \cr} $$ where the angle brackets denotes the Eulerian N. of 1st kind