Determine the Laurent Series for $f(z)= \frac{e^z}{z-1} , |z| > 1 $ This is a problem from an introductory course in complex analysis for engineers.
$f$ is analytical everywhere where $z \neq 1$ but since $|z| > 1$ this is already taken care of. 
Wrong attempt: We want to find a geometric series that "fit" the function, right? If so, one way is to notice that:
\begin{equation}
f= -\frac{e^z}{1-z} = \frac{1}{z} \frac{e^z}{1 - \frac{1}{z}} = \frac{1}{z} \cdot e^z \frac{1}{1 - \frac{1}{z}} = \frac{1}{z} \cdot e^z \sum_{n=0}^{\infty} \frac{(-1)^n}{z^n} =  e^z \sum_{n=0}^{\infty} \frac{(-1)^n}{z^{n+1}}
\end{equation} 
and another way to write this, by starting the series at $n=1$, is
\begin{equation}
e^z \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{z^{n}}
\end{equation} 
But this is not the correct solution. Any help appreciated!
Idea: Am I supposed to multiply the series for $e^z$ and $\frac{1}{z-1} = (z-1)^{-1}$ and then use the fact that the laurent series is unique and identify the coefficients? 
Solution from my textbook: $ \sum_{n=1}^{\infty} \frac{e}{z^{n}} +  \sum_{n=0}^{\infty}( e -  \sum_{k=0}^{n} \frac{1}{k!}) z^n $

Truly : 
$$f(z)= \frac{e^z}{z-1} =\frac{e^z}{z(1-1/z)}$$
A common geometric series is : 
$$\frac{1}{1-w} =\sum_{n=0}^\infty w^n,|w|<1 $$
About the exponential function, it is : 
$$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$
So, that goes along as : 
$$f(z) = \frac{e^z}{z}\sum_{n=0}^\infty z^{-n}=\sum_{n=0}^\infty \frac{z^n}{n!}\sum_{n=0}^\infty z^{-(n+1)}$$
for $\big|\frac{1}{z}\big| < 1 \Leftrightarrow |z|>1$. Now, if you calculate the common sum of these two, you'll get your result (it's easy) !
On the other hand, if you would want the Laurent Series around $z=1$, you can develop the Laurent series (which is in fact a Taylor series) of $e^z$ about $z=-1$:
$$e^z = \sum_{n=0}^\infty \frac{e}{n!}(z-1)^n $$
because $\frac{d^n}{dz^n}e^z=e^z$. Now just divide: 
$$\frac{e^z}{z-1}= \frac{1}{z-1} \sum_{n=0}^\infty \frac{e}{n!}(z-1)^n=  \sum_{n=0}^\infty \frac{e}{n!}(z-1)^{n-1}$$
