Calculating residue of $f(z) = \frac{(z^2 - 1)^4}{z^5}$ at $z=0$ 
Calculate the residue of  $$f(z) = \frac{(z^2 - 1)^4}{z^5}$$ in $z = 0$

I let $g(z) = (z^2 - 1)^4.$
I'm using a theorem which states:

Suppose $g$ is holomorphic around $z = \alpha$ and that $N$ is a
  positive integer, then $$RES_{z=\alpha}\frac{g(z)}{(z-\alpha)^N} =
 \frac{g^{(N-1)}(\alpha)}{(N-1)!}$$

and I get the correct answer (since $g(z) / (z-0)^5 = f(z))$, however, it is really annoying to have to differentiate that many times. Is there a smarter method? 

$$
\frac{(z^2 - 1)^4}{z^5}
= \frac{z^8 - 4 z^6 + 6 z^4 - 4 z^2 + 1}{z^5}
= \cdots + \frac{6}{z} + \cdots
$$
