Derivative of the Gamma function How do you prove that
$$
\Gamma'(1)=-\gamma,
$$
where $\gamma$ is the Euler-Mascheroni constant?

The Weierstrass product for the $\Gamma$ function gives:
$$\Gamma(z+1)=e^{-\gamma z}\cdot\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}\tag{1}$$
hence by considering $\frac{d}{dz}\log(\cdot)$ of both terms we get:
$$ \psi(z+1)=\frac{\Gamma'(z+1)}{\Gamma(z+1)}=-\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right) \tag{2}$$
and by evaluating the previous identity in $z=0$ it follows that:
$$ \psi(1) = \Gamma'(1) = -\gamma.\tag{3}$$
