Using term-by-term differentiation, expand $\displaystyle f(z)=\frac{1}{(z+3)^2}$ in a Laurent series valid for $\displaystyle |z| > 3$.

Ok, so, suppose I get a few derivatives:

$\displaystyle f'(z)=-2(z+3)^{-3}$

$\displaystyle f''(z)=6(z+3)^{-4}$

$\displaystyle f'''(z)=-24(z+3)^{-5}$

...

There's infinitely many. What next?