Proof: We have $$\frac{1}{\Gamma(x)}=xe^{\gamma x}\prod_{n=1}^{\infty} \left(\left(1+\frac{x}{n}\right)e^{-x/n}\right)$$ and thus $$\frac{1}{\Gamma(x)}\frac{1}{\Gamma(-x)}=-x^2e^{\gamma x}e^{-\gamma x}\prod_{n=1}^{\infty} \left(\left(1+\frac{x}{n}\right)e^{-x/n}\right)\left(\left(1-\frac{x}{n}\right)e^{x/n}\right)=-x^2\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)$$ But $\Gamma(1-x)=-x\Gamma(-x)$ and thus $$\frac{1}{\Gamma(x)}\frac{1}{\Gamma(1-x)}=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)$$ Now, using the formula for $\sin x/x$ we have $$\sin(\pi x)=\pi x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)$$ so that $$\frac{1}{\Gamma(x)}\frac{1}{\Gamma(1-x)}=\frac{\sin(\pi x)}{\pi}$$ and the result follows.