Laurent Series $e^{\frac1{1-z}}$, $|z|>1$. I tried to find Laurent expansion for:
$e^{\frac1{1-z}}$, $|z|>1$.
I tried next: $\frac1{1-z}=-\sum_{n=1}^{\infty}\frac1{z^n}$,
then using $e^{\frac1{1-z}}=1+\frac1{1-z}+\frac{1}{{2!(1-z)}^2}+\frac{1}{{3!(1-z)}^3}+...$
Then I have   $e^{\frac1{1-z}}$ = $1$ + $(-\frac1{z}-\frac1{z^2}-\frac1{z^3}-...) + (\frac1{2!})(\frac1{z^2}+\frac1{z^4}+…)+...$ = $1 - \frac1z-\frac1{2z^2}-...$. 
The result is wrong, since in the book the answer is: $1 - \frac1z+ \frac1{2z^2}-\frac1{6z^3}+\frac1{24z^4}-\frac{19}{120z^5}...$
What I did wrong?
 A: Note that$$\left(-\frac1z-\frac1{z^2}-\frac1{z^3}-\frac1{z^4}-\cdots\right)^2=\frac1{z^2}+\frac2{z^3}+\frac3{z^4}+\cdots,$$that$$\left(-\frac1z-\frac1{z^2}-\frac1{z^3}-\frac1{z^4}-\cdots\right)^3=-\frac1{z^3}-\frac3{z^4}-\frac6{z^5}-\cdots,$$and so on. This explains why you got the wrong result.
A: With $z=\dfrac{1}{w}$ we have $|w|<1$ then by $e^z$ expansion
\begin{align}
e^{\frac{1}{1-z}}
&= e^{\frac{-w}{1-w}} \\
&= \sum_{n\geq0}\dfrac{1}{n!}\left(\frac{-w}{1-w}\right)^n \\
&= \sum_{n\geq0}\dfrac{(-w)^n}{n!}\left(1-w\right)^{-n} \\
&= \sum_{n\geq0}\dfrac{(-w)^n}{n!}\left(1+nw+\dfrac{n(n+1)}{2}w^2+\cdots\right) \\
&= \sum_{n\geq0}\dfrac{(-1)^n}{n!}\left(w^n+nw^{n+1}+\dfrac{n(n+1)}{2}w^{n+2}+\cdots\right)
\end{align}
after calculation some terms we let $w=\dfrac1z$.

Edit: One may find 
\begin{align}
&1 \\
&-(w+w^2+w^3+w^4+w^5+\cdots) \\
&+\frac12(w^2+2w^3+3w^4+4w^5+\cdots) \\
&-\frac16(w^3+3w^4+6w^5+\cdots) \\
&+\cdots\\
&=1-w-\frac{w^2}{2}-\frac{w^3}{6}+\frac{w^4}{24}+\frac{19 w^5}{120}+\cdots\\
&=\color{blue}{1-\dfrac1z-\frac{1}{2z^2}-\frac{1}{6z^3}+\frac{1}{24z^4}+\frac{19}{120z^5}+\cdots}
\end{align}
