I have also posted this question at http://math.stackexchange.com/questions/486917/simple-approximation-to-a-sum-involving-stirling-numbers. I have an exact answer to a problem, which is the function:

$f(x,y)=\frac{1}{y^x}\sum_{i=1}^{x-1}{[i\binom{y}{x-i}(x-i)!S(x,x-i)]}$ where $S(x,x-i)$ is Stirling number of the second kind. Equivalently, $f(x,y)=\frac{1}{y^x}\sum_{i=1}^{x-1}{\{i\binom{y}{x-i}\sum_{j=0}^{x-i}{[(-1)^{x-i-j}\binom{x-i}{j}j^x]}\}}$. Equivalently, $f(x,y)=\frac{y!}{y^x}\sum_{i=1}^{x-1}{\{\frac{i}{(y-(x-i))!}\sum_{j=0}^{x-i}{[\frac{(-1)^{x-i-j}j^x}{j!(x-i-j)!}]}\}}$.

I have noticed that the percent difference between $f(x,y)$ and $g(x,y)$ goes to $0$ for larger values of $x$ and $y$, where $g(x,y)$ is the far more elegant $x-y(1-e^{-\frac{x}{y}})$. How can $f(x,y)$ be approximated by $g(x,y)$? What approximations should be used to make this connection?

I have tried approximations for $S(n,m)$ listed at http://dlmf.nist.gov/26.8#vii, to no avail.