What is the Laurent series expansion of $e^{-e^{\frac {1} {z}}}$ about $z=0$? 
What is the Laurent series expansion of $e^{-e^{\frac {1} {z}}}$ about $z=0$?

I know that if the Laurent series expansion of a function $f$ about $z=0$ is 
$$f(z)= \cdots + \frac {b_2} {z^2} + \frac {b_1} {z} + b_0 + b_1 z + b_2 z^2 + \cdots$$
then $$b_k = \oint_{\gamma} \frac {f(z)} {z^{k+1}}\ dz, \ k \in \Bbb Z.$$ where $\gamma$ is the unit circle centered at the origin. 
I also know that if $f$ is entire and if $$f(z)=\sum_{k=0}^{\infty} a_k z^k$$ be the Taylor's series expansion of $f$ about $z=0$ then the Laurent's series expansion of $g(z)=f(\frac {1} {z})$ (which is analytic on $\Bbb C \setminus \{0 \}$) about $z=0$ is $$g(z) = \sum_{k=0}^{\infty} \frac {a_k} {z^k}.$$
Using these facts I have tried to find out the Laurent series expansion for $e^{-e^{\frac {1} {z}}}$ about $z=0$. Clearly in this case the Laurent's series expnsion for $e^{\frac {1} {z}}$ about $z=0$ is $$e^{\frac {1} {z}} = \sum_{k=0}^{\infty} \frac {1} {k! z^k}$$ since  we know that Taylor's series expansion of $e^z$ about $z=0$ is $$e^z=\sum_{k=0}^{\infty} \frac {z^k} {k!}.$$ 
From the above discussion can I conclude that the Laurent's series expansion of $e^{-e^{\frac {1} {z}}}$ about $z=0$ is $$e^{-e^{\frac {1} {z}}} = \sum_{j=0}^{\infty} (-1)^{j}\frac {1} {j!} \left ( \sum_{k=0}^{\infty} \frac {1} {k!z^k} \right )^{j}?$$
If the answer is 'yes' then why? Also I want to know what will be the residue of $e^{-e^{\frac {1} {z}}}$ at $z=0$ where I am really struggling. Please help me in this regard.
Thank you very much.
 A: As you have said that if we know the Taylor series expansion of an entire function $f$ about $z=0$ as lets say $$f(z)=\sum_{k=0}^{\infty} a_k z^k$$ then the Laurent series expansion of $g(z)=f(\frac 1 z)$ about $z=0$ is $$g(z)=\sum_{k=0}^{\infty} \frac {a_k} {z^k}.$$ We use this fact to evaluate the Laurent series expansion of $e^{-e^{\frac {1} {z}}}$ about $z=0$. Let us first find the Laurent series expansion of $f(z)=e^{-e^z}$ about $z=0$. Clearly $f$ thus defined is entire. Hence we can find the Taylor series expansion of $f$ about $z=0$ as follows $:$
$$f(z)=\sum_{k=1}^{\infty} \frac {f^{(k)}(0)} {k!} z^k.$$ where $f^{(k)}(0)$ denotes the $k$-th order derivative of $f$ at $z=0$. If you evaluate the higher order derivatives of $f$ at $z=0$ then you will find that 
$$
f^{(k)}(0) = \left\{
        \begin{array}{ll}
            \frac {1} {e} & \quad k = 3n,\ \ \ n \in \Bbb Z^+ \cup \{0 \}\\
            {-\frac {1} {e}} & \quad k=3n-2,\ \ \ n \in \Bbb Z^+\\\
            0 & \quad k=3n-1,\ \ \ n \in \Bbb Z^+
        \end{array}
    \right.
$$
Hence the Taylor series expansion of $f$ about $z=0$ is $$f(z)=\frac {1} {e}\left (1-z+\frac {z^3} {3!}- \frac {z^4} {4!} + \cdots  \right ).$$
Therefore the required Laurent series expansion of $e^{-e^{\frac {1} {z}}}$ about $z=0$ is same as the Laurent series expansion of $g(z)=f(\frac {1} {z})$ about $z=0$ which is $$g(z)=\frac {1} {e} \left (1-\frac {1} {z} + \frac {1} {3! z^3} - \frac {1} {4! z^4} + \cdots \right ).$$
From here you can easily see that the residue of $e^{-e^{\frac {1} {z}}}$ at $z=0$ is $-\frac {1} {e}$.
A: You are on the right track, but that is not the Laurent expansion (the powers of $z$ are not isolated).
\begin{align}
e^{-e^{1/z}}&=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\,e^{n/z}\\
&=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\sum_{j=0}^\infty\frac{1}{j!}\frac{n^j}{z^j}\\
&=\sum_{j=0}^\infty\frac{1}{j!}\Bigl(\sum_{n=0}^\infty\frac{(-1)^nn^j}{n!}\Bigr)\frac{1}{z^j}.
\end{align}
The residue at $z=0$ is the coefficient of $1/z$ ($j=1$), which is
$$
\sum_{n=0}^\infty\frac{(-1)^nn}{n!}=\sum_{n=1}^\infty\frac{(-1)^n}{(n-1)!}=-\sum_{n=0}^\infty\frac{(-1)^n}{n!}=-\frac1e.
$$
