Rewrite $\Gamma(-z)$ in terms of $\Gamma(z)$

There is the Euler's reflection formula:, which, together with the functional equation for Gamma ($\Gamma(t+1) = t \Gamma(t)$) gives you want you want.

We have that $\Gamma(1+z)=z\Gamma(z)$, or $\Gamma(1-z)=-z\Gamma(-z)$.
We also have that $\Gamma(1-z)\Gamma(z)=\dfrac{\pi}{\sin(\pi z)}$.
Putting these two together gives us that $$\Gamma(z)\Gamma(-z)=-\dfrac{\pi}{z\sin(\pi z)}$$