$$\frac{\sin(\pi x)}{\pi}=x\prod_{r=1}^\infty\left(1-\frac{x^2}{r^2}\right)$$ $$\Gamma(x)=\frac{1}{x}e^{-\gamma x}\prod_{r=1}^\infty\left(\frac{r}{x+r}\right)e^{\frac{x}{r}}$$ $$\Gamma(x) \Gamma(1-x)=\frac{1}{x(1-x)}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{r^2}{(x+r)(1-x+r)}\right)e^{\frac{1}{r}}$$ Assume $\Gamma(x) \Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$, then, for $x \notin \mathbb{Z}$:
$$\frac{1}{x}\prod_{r=1}^\infty\left(\frac{r^2}{r^2-x^2}\right)=\frac{1}{x(1-x)}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{r^2}{(x+r)(1-x+r)}\right)e^{\frac{1}{r}}$$ $$\prod_{r=1}^\infty\left(\frac{1}{r-x}\right)=\frac{1}{1-x}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{1}{1-x+r}\right)e^{\frac{1}{r}}$$ $$\prod_{r=1}^\infty\left(\frac{1}{r-x}\right)=\frac{1}{1-x}e^{-\gamma}\prod_{r=1}^\infty e^{\frac{1}{r}}\prod_{r=2}^\infty\left(\frac{1}{r-x}\right)$$ $$1=e^{-\gamma}\prod_{r=1}^\infty e^{\frac{1}{r}}$$ Thus $-\gamma+\sum_{r=1}^\infty \frac{1}{r}=0$, which is ridiculous. Where did I go wrong?
In the line
$$\prod_{r=1}^\infty\left(\frac{1}{r-x}\right)=\frac{1}{1-x}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{1}{1-x+r}\right)e^{\frac{1}{r}}$$
you have divergent products. If one wants to give meaning to it, the only somewhat sensible meaning is $0 = 0$. Of course dividing that equation by $0$ does not produce anything meaningful.