Euler's reflection formula

Step 1 (Reduce to the Beta function). Using $B(x,y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ with $y=1-z$, $\Gamma(z)\Gamma(1-z) = B(z,1-z) = \int_0^1 t^{z-1}(1-t)^{-z}\,dt$. Substituting $u = t/(1-t)$ turns this into $\displaystyle\int_0^\infty \dfrac{u^{z-1}}{1+u}\,du$ for $0<\operatorname{Re}(z)<1$.

Step 2 (Set up a keyhole contour). To evaluate $I=\int_0^\infty \dfrac{u^{z-1}}{1+u}\,du$, use the branch of $w^{z-1}$ cut along the positive real axis and integrate $\dfrac{w^{z-1}}{1+w}$ around a keyhole contour: outward just above the cut, a large circle of radius $R$, back just below the cut, and a small circle of radius $\varepsilon$ around $0$.

Step 3 (Residue at the only pole). The only singularity of $\dfrac{w^{z-1}}{1+w}$ inside the contour is the simple pole at $w=-1$, where (using the branch with argument $\pi$ there) $\mathrm{Res} = (-1)^{z-1} = e^{i\pi(z-1)}$, so the residue theorem gives the keyhole integral as $2\pi i\, e^{i\pi(z-1)}$.

Step 4 (Compare the two sides of the cut and let $R\to\infty$, $\varepsilon\to 0$). The large and small circles vanish for $0<\operatorname{Re}(z)<1$, and the two straight segments differ by the phase picked up by $w^{z-1}$ crossing the cut, giving $I\big(1 - e^{2\pi i(z-1)}\big) = 2\pi i\, e^{i\pi(z-1)}$. Solving for $I$ and simplifying the exponentials using $\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$ gives $I = \dfrac{\pi}{\sin(\pi z)}$, so combining with Step 1, $\Gamma(z)\Gamma(1-z) = \dfrac{\pi}{\sin(\pi z)}$.