This seems like it must have been addressed somewhere already, but I cannot find it in any standard series tables.


I have the equation:


$f(z) = \left(1 + \frac{1}{z}\right)^z$. 


What is the general form for the $n$th term of the series? That is, if I have


$f(z) \sim \sum\_{n=0}^{\infty} \frac{c\_n}{z^n}$


near $z = \infty$, what is the form of $c\_n$?



What is the series expression for (1+1/x)^x about x = \infty?
Markus Brede proves the following formula in the paper ["On the convergence of the sequence defining Euler’s number"](http://www.springerlink.com/content/c1q27x0730v30hth/). Let $$\left(1+\frac{1}{z}\right)^z=\sum\_{n\geq 0} \frac{a\_n}{z^n}$$
then we have
$$a\_n=e\sum\_{v=0}^n \frac{S(n+v,v)}{(n+v)!}\sum\_{m=0}^{n-v}\frac{(-1)^m}{m!}$$
where $S(a,b)$ are Stirling numbers of the first kind. This shows that all coefficients are rational multiples of $e$. I found the article through [OEIS](http://oeis.org/search?q=1%2C11%2C7%2C2447&sort=&language=english&go=Search).


