Can $\sum_{i=0}^n (k^i)/i!$ be simplified? Can this expression be simplified?
$ \sum_{i=0}^n (k^i)/i!  $
EDIT:
This expression I have got as the number of possible ways to select n objects or less from k different infinite objects (you can select as many as you can from any type)...
I believe it must be equal to the number of possible ways to select n objects from k+1 different infinite objects, where the extra +1 is a dummy object, selecting one of this type means letting an empty space in the final selection (i.e. selecting n-1 not n)...
so this must be equal to:
$ (k+1)^n/n! $
Why that is not right??
 A: Your reasoning is incorrect. $\frac{k^i}{i!}$ is not even an integer in general, so it can't count what you're claiming it counts. While $k^i$ is the number of ways to choose $i$ objects from $k$ types of objects in order, you can't just divide by $i!$ to get rid of the ordering because some of the orderings are the same when you rearrange them.
For example, if $k = 2, i = 3$, then the possible ways to choose $3$ objects from $2$ types of objects in order are $AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB$, and the possible ways to choose $3$ objects from $2$ types are $AAA, AAB, ABB, BBB$, which is $4$ and not $\frac{2^3}{3!} = \frac{4}{3}$. When you are dividing by $3!$ you are pretending that the different orderings group evenly into groups of $6$ but, as you can see, they don't.
The actual number you want, for fixed $i$, is the number of multisets of $i$ objects on a $k$-element set, which is actually
$${k+i-1 \choose i}.$$
This can be proven using what's called the "stars and bars" argument, as well as Burnside's lemma. Note that if $i$ is fixed and $k$ tends to infinity this is asymptotically, but not exactly, $\frac{k^i}{i!}$.
As you say, if you want the number of multisets of $n$ or less objects, then you just add in a dummy element of the set, so the final answer is
$$\sum_{i=0}^n {k+i-1 \choose i} = {k+n \choose n}.$$
This identity is sometimes known as the hockey-stick identity because it has an interpretation in terms of summing along a hockey-stick-like shape on Pascal's triangle. 
A: In response to the edited question, it is not right.  If we just let n=2, $1+k+\frac{k^2}{2}\neq \frac{(k+1)^2}{2}$.  The difference is $\frac{1}{2}$
A: Your existing formula does not produce integers all the time:
Consider number of types, k=1
I believe that means f(k,n)=n
Your f(1,n)=
$ \sum_{i=0}^n 1/i!  $
 which is definitely not generally an integer.  
What you want is  
$ \sum_{i=0}^n g(k,i)  $  
where g(k,i) is the number of ways you can select exactly i objects from upto k types.  
Which can be similarly expanded to  
$ \sum_{i=0}^n \sum_{j=0}^k h(j,i)  $  
where h(j,i) is the number of ways you can select exactly i objects from exactly j types.  
Note, because h(j,i)=0,i<j, we can swap the summations:
$ \sum_{j=0}^k \sum_{i=j}^n  h(j,i)  $  
[  and l(j,n)=$ \sum_{i=j}^n  h(j,i)  $ is the number of ways you can select between j and n objects from exactly j types  

which is the same as selecting between 0 and n-j objects from up to j types  

thus l(j,n)=$ \sum_{i=0}^{n-j}  g(j,i)  $  

so f(k,n)=$ \sum_{i=0}^n g(k,i) =  \sum_{j=0}^k \sum_{i=0}^{n-j}  g(j,i)  $  ]  
One of g or h (or l) should be definable using combinations as suggested by Yuval Filmus's comment.
