Find the residue, state the nature of the singularity, find the constant term in $1/\sin(ze^z)$ at $z=0$ 
Find the residue, state the nature of the singularity, find the constant term in series $1/\sin(ze^z)$ at $z=0$.

We can rewrite the function $\frac{1}{\sin(ze^z)}$ as $\frac{ze^z}{\sin(ze^z)}\cdot\frac{1}{ze^z}$. What I did next I think might not be true and that's why I'm writing this post.
Since $\lim\limits_{w\rightarrow0}\frac{w}{\sin w}=1$ and $\lim\limits_{z\rightarrow 0}ze^z=0$ and $[{d\over dz}e^z]_{z=0}=1$ basically I said that at zero we can just look at $\frac{1}{ze^z}={1\over z}+\sum\limits_{n=1}^\infty\frac{(-1)^n}{n!}z^{n-1}$ as the Laurent series of our original function.
But I feel like this is far from rigorous... 
From this it results that the residue at zero is $1$, which is true for the original function;
the constant term is $-1$, also true for $\frac{1}{\sin(ze^z)}$ and $z_0=0$ is a pole of degree $1$ also true.
 A: The function $g(z)=ze^z$ is analytic and $g'(z)=e^z+ze^z$ is different from $0$ in a neighborhood of $0$, so it is invertible on that neighborhood, with analytic inverse.
Thus the substitution $w=ze^z$ is possible in a limit and
$$
\lim_{z\to0}\frac{z}{\sin(ze^{z})}=
\lim_{z\to0}\frac{ze^z}{\sin(ze^{z})}e^{-z}=
\lim_{w\to0}\frac{w}{\sin w}e^{-g^{-1}(w)}=1
$$
Therefore $f(z)=1/\!\sin(ze^{z})$ has a pole of order $1$ at $0$ and the residue is $1$.
The derivative of $zf(z)$ (for $z\ne0$) is
$$
\frac{\sin(ze^z)-z(e^z+ze^z)\cos(ze^z)}{\sin^2(ze^z)}=
\frac{\sin w-w\cos w-g^{-1}(w)w\cos w}{\sin^2w}
$$
Note that $g^{-1}(w)=w+o(w)$, because its derivative at $0$ is $1$, so we have
$$
\lim_{z\to0}\frac{\sin(ze^z)-z(e^z+ze^z)\cos(ze^z)}{\sin^2(ze^z)}=
\lim_{w\to0}\frac{w-w-w^2+o(w^2)}{w^2+o(w^2)}=-1
$$
Thus
$$
zf(z)=1-z+o(z^2)
$$
and finally
$$
f(z)=\frac{1}{z}-1+o(z)
$$
A: With the help of a computation package, I found that, at $z=0$,
$$
\sin(z\exp(z)) = z + z^2 + \frac13z^3 - \frac13z^4 - \frac7{10}z^5 + \cdots\,,
$$
and its reciprocal is
$$
\frac1{\sin(z\exp(z))} =
z^{-1} - 1 + \frac23z + \frac{13}{90}z^3 + \frac7{90}z^4 + \frac{37}{378}z^5 +\cdots\,.
$$
You could certainly determine the first few terms of these by mindless hand-computation.
I think, though, that seeing that the residue is $1$ is just a matter of direct examination without any kind of computation at all.
A: You can find the first few terms of the series, step-by-step
$$ ze^z = z\left(1 + z + \frac{z^2}{2} + \dots \right) = z + z^2 + \frac{z^3}{2} + \dots $$
\begin{align} \sin(ze^z) &= \sin\left(z + z^2 + \frac{z^3}{2} + \dots \right) \\
&= \left(z + z^2 + \frac{z^3}{2} \dots \right) - \frac{1}{3!}\left(z + z^2 + \frac{z^3}{2} +\dots \right)^3 + \dots \\
&= z + z^2 + \frac{z^3}{2} - \frac{z^3}{6} + \dots \\
&= z + z^2 + \frac{z^3}{3} + \dots \end{align}
\begin{align} 
\frac{1}{\sin(ze^z)} &= \frac{1}{z+z^2+\frac{z^3}{3}+\dots} \\
&= \frac{1}{z} \frac{1}{1 + z + \frac{z^2}{3} + \dots} \\
&= \frac{1}{z} \left[1 - \left(z + \frac{z^2}{3} + \dots\right) + \left(z + \frac{z^2}{3} + \dots\right)^2 - \dots \right] \\
&= \frac{1}{z}\left[1 - z - \frac{z^2}{3} + z^2  + \dots\right] \\
&= \frac{1}{z}\left[1 - z + \frac{2z^2}{3} + \dots \right]
\end{align}
so the residue is $1$ and the constant term is $-1$
