Laurent expansion of 1/sin(z)

Question

I've been looking for a concise explanation of how to obtain the Laurent expansion for $$\frac{1}{\sin(z)}$$ My attempt at it has me confused by it pretty quickly. I start with the knowledge that $$\sin(z)=z-\frac{1}{3!}z^3+\frac{1}{5!}z^5-...$$ Then I move that to the denominator to begin expanding the function at hand. I have gotten it to the form of a geometric series; namely, $$\frac{1}{\sin(z)}=\frac{1}{z}\cdot\frac{1}{1-(\frac{1}{3!}z^2-\frac{1}{5!}z^4+\cdots)}$$ I am unsure of how to proceed from here. Any help appreciated.