Calculate the poles and singularity of $f(z)=\frac{z\sin z}{(1-e^z)^3}$

Question

Calculate the poles and classify the singularity of $f(z)=\frac{z\sin z}{(1-e^z)^3}$
My attempt:

I know that $f(z)$ have a singularity in $z_0=0$.

I don't know how see the order of the pole, but i know $z_0$ is a pole if

$$\lim_{z\rightarrow z_0} (z-z_0)^nf(z)$$ exists and not equal to 0, then:

$\lim_{z\rightarrow z_0} (z-z_0)^nf(z)=\lim_{z\rightarrow 0}z^n\frac{z\sin z}{(1-e^z)^3}$

here i'm stuck.

Answer 1

Since $z\sin z$ has a double zero at $0$, whereas $(1-e^z)^3$ has a triple zero there, $\frac{z\sin z}{(1-e^z)^3}$ has a simple pole there ($2-3=-1$).