The formula for $\Gamma$ function which is valid for all complex values of the argument was introduced by Euler, $$\Gamma(z)=\frac{1}{z}\prod_{n=1}^\infty\left\{\left(1+\frac{1}{n}\right)^z\left(1+\frac{z}{n}\right)^{-1}\right\}.$$ This is how he originally defined it, and then he showed that it is equal to the "Euler's integral". Of course at the time of Euler there was no rigorously defined notion of analytic continuation, but this did not prevent Euler from working with it.

A modern definition using the product (which also defines $\Gamma$ for all complex $z$) is due to Weierstrass: $$\frac{1}{\Gamma(z)}=ze^{\gamma z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-z/m},$$ where $\gamma$ is the Euler-Mascheroni constant.

Source: Whittaker-Watson, Chapter 12.

The formula for $\Gamma$ function which is valid for all complex values of the argument was introduced by Euler, $$\Gamma(z)=\frac{1}{z}\prod_{n=1}^\infty\left\{\left(1+\frac{1}{n}\right)^z\left(1+\frac{z}{n}\right)^{-1}\right\}.$$ This is how he originally defined it, and then he showed that it is equal to the "Euler's integral". Of course at the time of Euler there was no rigorously defined notion of analytic continuation, but this did not prevent Euler from working with it.

A modern definition using the product (which also defines $\Gamma$ for all complex $z$) is due to Weierstrass: $$\frac{1}{\Gamma(z)}=ze^{\gamma z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-z/n},$$ where $\gamma$ is the Euler-Mascheroni constant.

Source: Whittaker-Watson, Chapter 12.