Show $e^x / (1 - x)^n$ is the exponential generating function for a specified sequence 
Show that $e^x/(1-x)^n$ is the exponential generating function for the number of ways to choose some subset (possibly empty) of $r$ distinct objects and distribute them into $n$ different boxes with the order in each box counted. 

I know I have to use:
$f(x)=e^x=\sum x^r/r!$ and $g(x)=1/(1-x)^n= \sum C(r+n-1,r)x^r$
I'm not sure where to go from here.

By way of enrichment, here is the species specification for this problem:
$$\mathfrak{P}(\mathcal{Z}) \times 
\mathfrak{S}_{=n}(\mathfrak{S}(\mathcal{Z})).$$
The first term represents the set of values that are exempted from being distributed into the $n$ boxes and the rest describes ordered partitions being distributed into $n$ slots.
These ordered partitions are sequences which we use because the order in each box matters.
Translating to generating functions we immediately obtain
$$\exp(z)\times \left(\frac{1}{1-z}\right)^n.$$
E.g. there are $n!$ different sequences on $n$ labelled objects, giving the EGF operator
$$\sum_{n\ge 0} n! \frac{z^n}{n!} = \frac{1}{1-z}.$$
Alternatively, the group $E_n$ containing just the identity acting on $n$ objects consists of just one permutation, producing $n!/|E_n|$ orbits, giving again
$$\sum_{n\ge 0} \frac{n!}{|E_n|} \frac{z^n}{n!} = \frac{1}{1-z}.$$
For an explanation of these operators there is the 
Wikipedia entry for symbolic combinatorics.
