Challenging Laurent series

I want to find the Laurent series expansion of $f(z)=\frac{e^{2z}}{z^2}$, and apprently, the pole is at 0, so this series would be for $|z|>0$.

I am no sure this is correct, but if I consider the numerator, it is in the form of $e^w$, which has the Laurent expansion as $\frac{w^n}{n!}$, so a nice suggestion would be

\begin{equation} f(z)=\frac{\lim_{n\longrightarrow \infty}\frac{(2z)^n}{n!}}{z^2} \end{equation}

But this appears too simple, and must be wrong. I couldn't find anything similar on Stackexchange. Are there any suggestions?

Thanks

Since

$$e^{2z} = 1 + \frac{2z}{1!} + \frac{4z^2}{2!} + \frac{9z^3}{3!} + \frac{16z^4}{4!} +\cdots$$

just divide both sides by $z^2$.