How may one go about proving
$\displaystyle\frac{\Gamma'(s)}{\Gamma(s)}=O(\log|s|)$,
(away from the poles) directly? By a direct proof, I mean not to go through the usual Stirling formula with the exact error term. The use of a rough form of Stirling's formula is welcome.
You can use the product representation of $\Gamma(z)$, take logs and differentiate the resulting series.
$$\Gamma(z) = \dfrac{e^{-\gamma z}}{z} \prod_{n=1}^{\infty} \left(1 + \dfrac{z}{n}\right)^{-1} \ e^{\frac{z}{n}}$$
You can find more information here: Polygamma Function and Digamma Function.