Derivative of the Gamma function

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

Consider the integral form of the Gamma function, \begin{align} \Gamma(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, dt \end{align} taking the derivative with respect to $x$ yields \begin{align} \Gamma'(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, \ln(t) \, dt. \end{align} Setting $x=1$ leads to \begin{align} \Gamma'(1) = \int_{0}^{\infty} e^{-t} \, \ln(t) \, dt. \end{align} This is one of the many definitions of the Euler-Mascheroni constant. Hence, \begin{align} \Gamma'(1) = - \gamma = \int_{0}^{\infty} e^{-t} \, \ln(t) \, dt. \end{align}

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: $$ \psi(1) = \Gamma'(1) = -\gamma.\tag{3}$$