about formulas and identies for Stirling numbers of the second kind How the two following formulas can be proved (algebraically preferred)?
$$\sum_{n=k}^{∞}S\left(n,k\right)\ \frac{x^{n}}{n!}=\frac{1}{k!}\left(e^{x}-1\right)^{k}$$
$$x^{n}=\sum_{m=0}^{n}S\left(n,m\right)\left(x\right)_{m}$$
where $S\left(n,m\right)$ is the Stirling number of the second kind and $\left(x\right)_{m}$ is falling 
factorial.
any Hint or full proof is highly appreciated (since I'm new to this numbers and their relations so a full 
proof is better).
 A: If  you are  familiar with  combinatorial classes  the first  equation
represents a  set of $k$ elements,  each of these themselves  a set of
labeled  elements, with  a non-zero  number of  elements (here  we are
distributing  $n$  distinguishable  balls into  $k$  indistinguishable
boxes with no box being empty):
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}_{=k}
(\textsc{SET}_{\ge 1}(\mathcal{Z})).$$
The         exponential         generating        function         for
$\textsc{SET}_{=k}(\mathcal{Z})$  is  $$\frac{z^k}{k!}.$$ This  yields
for set partitions into $k$ sets
$$\frac{1}{k!}
\left(\sum_{q=1}^\infty \frac{z^q}{q!}\right)^k
= \frac{1}{k!} (\exp(z)-1)^k.$$
This is because in a combinatorial class constructed by distributing a
repertoire of source objects into a row  of $k$ slots with a group $G$
permuting the slots  and creating equivalence classes, the  EGF of the
class is given by
$$\frac{S(z)^k}{|G|}.$$
where $S(z)$ is  the EGF of the  source objects.  For sets  $G$ is the
symmetric  group  and $|G|=k!.$  These  combinatorial  classes may  be
nested.  Another example  are cycles  with  $|G|=k$ where  we get  the
generating function
$$\sum_{q\ge 1} \frac{z^q}{q} = \log\frac{1}{1-z}$$
for the class of cycles. Of course permutations are sets of cycles
and we have
$$n! [z^n] \exp  \log\frac{1}{1-z} = n! [z^n] \frac{1}{1-z}
= n!.$$
For the  second equation we  have the following  claim where $x$  is a
positive integer:
$$x^n = \sum_{m=0}^n {n\brace m} x^{\underline m}
= \sum_{m=1}^n  {x\choose m} m! {n\brace m}.$$
Suppose we throw  $n$ different balls into $x$  different boxes, there
are $x^n$ ways of doing this.  On the other hand we may classify every
distribution of balls obtained in this  way by the number $m$ of boxes
that were not  empty.  To get this kind of  distribution we choose the
$m$ boxes in ${x\choose m}$ ways  and partition the $n$ balls into $m$
sets  in ${n\brace  m}$ ways.  These $m$  sets can  be matched  to the
chosen $m$ boxes in $m!$ ways and every such configuration constitutes
a distribution of the balls, and we have equality.
 Given  that LHS and  RHS may be viewed  as polynomials in  $x$ and
they  are equal  at all  positive integer  values, they  are the  same
polyonmial.
