Show that $\Gamma(a -S) = \frac{\Gamma(1-a+S)\Gamma(a)}{(-1)^S\Gamma(a-S)}$

Question

If $S$ is an integer and $a$ is a fractional, then $\Gamma(a -S) = \frac{\Gamma(1-a+S)\Gamma(a)}{(-1)^S\Gamma(a-S)}$.

I think that it can be used $\Gamma(x)\Gamma(1-x) = \Gamma(a-S)\Gamma(1-a+S)=(a-S-1)!(S-a)!$, developing the factorials and using identity $\Gamma(z)=\frac{1}{z}\Gamma(z+1)$.

However, the minus sign of S gives me trouble. I don't know if I'm on the right track.