I think you can also use the following:
$\Gamma$ can also be expressed as
$\Gamma(z) = \frac{\exp{(-\gamma z)}}{z}\prod\limits_{n=1}^\infty\frac{\exp \left({\frac z n}\right)}{1+\dfrac z n }$
So that
$\log \Gamma(z)=-\gamma z-\log z+\sum\limits_{n=1}^\infty\frac{z}{n}- \sum\limits_{n=1}^\infty\log \left(1+\frac z n \right)$
Now I'm not going to address convergence now (which you can look up in Landau's Calculus, on the chapter dedicated to the Gamma function), but differentiating gives
\frac{\Gamma '(z)}{\Gamma(z)}=-\gamma-\frac 1 z+\sum\limits_{n=1}^\infty \left(\frac 1 n-\frac{1}{n+z}\right)
\frac{\Gamma '(z)}{\Gamma(z)}=-\gamma-\frac 1 z+\sum\limits_{n=1}^\infty\frac{z}{n(n+z)}
Letting $z=1$ gives:
\frac{\Gamma '(1)}{\Gamma(1)}=\Gamma'(1)=-\gamma-1 +\sum\limits_{n=1}^\infty\frac{1}{n(n+1)}
But we know that
$\sum\limits_{n=1}^\infty\frac{1}{n(n+1)}=1$
so that
\frac{\Gamma '(1)}{\Gamma(1)}=\Gamma'(1)=-\gamma-1+1=-\gamma