What will be the Laurent's series expansion of $e^{f(z)}$ about $z=0$? 
Let $f$ be a meromorphic function having a pole at $z=0$. Then what can we say about  the Laurent's series expansion of $e^f$ about $z=0$?

My confusion is in the following lines
"Can we first write $$e^{f(z)}=\sum\limits_{k=0}^{\infty} \frac {\{f(z) \}^{k}} {k!}$$ as a Taylor's series expansion of $e^f$ about $z=0$ and then take the Laurent's series for each $\{f(z) \}^{k}$ about $z=0$ to obtain the required Laurent's series about $z=0$?"
If the answer to this question is affirmative one then what is the reason behind it? Please help me in this regard.
Thank you very much.
 A: You'll want a suitable estimate to make sure that converges.  Suppose we take a circle $\Gamma$ of radius $r$ around $0$ so that all other singularities of $f$ are outside $\Gamma$.  We can then get the Laurent series coefficients $c_n$ of $e^f$ at $0$ by the generalized Cauchy integral formula
$$ c_n = \frac{1}{2\pi i} \oint_\Gamma \frac{e^{f(z)}}{z^{n+1}}\; dz $$
If $f$ has a pole of order $m$, the Laurent series for $f(z)$ is 
$$ f(z) = \sum_{j=-m}^\infty a_j z^j $$
converging absolutely and uniformly in a neighbourhood of $\Gamma$, thus there are positive constants $A$ and $B$ such that $|a_j| \le A B^j$ with $B r < 1$.  Now for $z \in \Gamma$,
$$ e^{f(z)} = \sum_{k=0}^\infty \frac{1}{k!} \left(\sum_{j=-m}^\infty a_j z^j\right)^k $$
The expansion of this is term-by-term bounded in absolute value by the corresponding terms of
$$ \sum_{k=0}^\infty \frac{1}{k!} \left( \sum_{j=-m}^\infty A B^j r^j \right)^k
= \sum_{k=0}^\infty \frac{1}{k!} \left(\frac{A B^{-m} r^{-m}}{1-Br}\right)^k
= \exp\left(\frac{A B^{-m} r^{-m}}{1-Br}\right)$$
which converges absolutely.  
So yes, we can re-arrange the series and the expansion works.
