What am I doing wrong while finding the Laurent series of $\frac{1}{\sin(z)}$? I'm trying to compute the Laurent series of $\frac{1}{\sin(z)}$ at $z_0=0$.
From what I've seen on the internet this is given as 
$f(z)=\frac{1}{z}+\frac{z}{3!}+\frac{7z^3}{360}+...$
My attempt:
$f(z)=\frac{1}{\sin(z)}$ can be rewritten as $f(z)=\frac{1}{z}\frac{z}{\sin(z)}$
We can express $\sin(z)$ as $\sin(z)=z-\frac{z^2}{3!}+\frac{z^4}{5!}+...$
$\therefore \frac{z}{\sin(z)}=(\frac{\sin(z)}{z})^{-1}=(1-\frac{z}{3!}+\frac{z^3}{5!}+...)^{-1}$
To get this to equal what I know it should, then we should have $(\frac{\sin(z)}{z})^{-1}=(1+\frac{z}{3!}+\frac{7z^3}{360}+...)$ but how can this be what do I do with my negative sign and how do I obtain the $\frac{7}{360}$ coefficient?

Note the most direct way to obtain  the expansion  of $\dfrac1{\sin z}$ is division by increasing powers. I'll show how to obtain the first  four terms:
Start from  the expansion of $\sin z$ at order $7$:$$\frac1{\sin z}=\frac1{z-\dfrac{z^3}{6}+\dfrac{z^5}{120}-\dfrac{z^7}{5040}+o(z^7)}=\frac1z\cdot\frac1{1-\dfrac{z^2}{6}+\dfrac{z^4}{120}-\dfrac{z^6}{5040}+o(z^6)}$$
Now the division of $1$ by $1-\dfrac{z^2}{6}+\dfrac{z^4}{120}-\dfrac{z^6}{5040}$ goes as follows (we truncate the terms  with degree more than $6$) :

Thus we obtain
\begin{align}
\frac1{\sin z} &= \frac1z \Bigl( 1 + \dfrac{z^2}{6} + \frac{7z^4}{360}-  \frac{z^6}{4320} +o(z^6)\Bigr) \\[1ex]
&=\frac1z + \dfrac{z}{6} + \frac{7z^3}{360}-  \frac{z^5}{4320} +o(z^5)
\end{align}
