Why doesn't the functional equation imply that $\zeta(s)=0$ for positive even integers?

Question

The Riemann Zeta Function satisfies the functional equation $\zeta(s)=2^s\pi^{s-1}\sin\left(\dfrac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$. But when $s$ is a positive even number, $\sin\left(\dfrac{\pi s}{2}\right)$ and $\Gamma(1-s)$ are both $0$, which would imply that $\zeta(2)=0$, but of course, $\zeta(2)=\dfrac{\pi^2}{6}$. What am I missing?

Answer 1

$\Gamma$ has a pole, not a zero, at each negative integer.