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.

Consider an exponential generating function $F(x)$.  The coefficient of $\frac{x^r}{r!}$ represents the number of ways to put some structure on $r$ labeled objects.
If $F$ and $G$ are two exponential generating functions, then the coefficient of $\frac{x^r}{r!}$ in $F(x) G(x)$ conveniently counts the number of ways to split $r$ labeled objects into two groups, then to put the structure counted by $F$ on the first group and structure counted by $G$ on the second group.
In this case, you need to split $r$ objects into a first group of things that are not put into any box, and a second group of things that are put into the $n$ boxes.  You should find that $e^x$ counts the first structure and $\frac{1}{(1-x)^n}$ counts the second.
To see that the coefficient of $\frac{x^r}{r!}$ in $\frac{1}{(1-x)^n}$ counts the number of ways to put $r$ objects into $n$ boxes, with the order in each box mattering, you can apply the same trick.  To put $r$ labeled objects in boxes in this way, you can first split the $r$ objects into $n$ groups, and then for each group count how many ways the objects can be put in the box.  There are $k!$ ways to put $k$ objects into a single box, so the generating function counting this is $\sum_{k=0}^\infty k! \frac{x^k}{k!} = \frac{1}{1-x}$.  Multiply this with itself $n$ times to get the function for $n$ boxes.
