I am given a function $f: G\rightarrow\mathbb{C}$, where $f\in\mathcal{H}(G)$, and where $f$ is not identically zero. Now if we assume that $f(a)=0$, we know that we can put $f$ on the form $$f(z)=(z-a)^ng(z)$$ where $g(a)\neq0$. My goal is now to find the number $n$ above, for the function $$f(z)=(e^z-1)^{12}\sin{z}$$ where we have $a=0$. My initial observation was that we can rewrite $$f(z)=(e^z-e^0)^{12}\sin{z}$$ and taking the principal log on both sides we get $$\log(f(z))=12\log(e^z-e^0)+\log(\sin{z})$$ However this seems to take me nowhere, i felt i was going the right way, since i basically want the $z-0$ down from the exponents.
This is not a homework question, but a preperation before my exam, any help is greatly appreciated
From the hint @Gary gave, by using that $$f(z)=z^{13}\left(\frac{e^z-1}{z}\right)^{12}\frac{\sin{z}}{z}$$ we know that $\sin{z}/{z}$ has a removable singularity at $z\rightarrow 0$, and so does $\left(\frac{e^z-1}{z}\right)^{12}$, this means that their product form $g(z)$, and thus $n=13$