$\gamma_r$ is the part of $|z-1|=r$ inside $|z|=1$, $\gamma_1$ is the part of |z|=1 outside $|z-1|<r$. prove that
1) If $|z|<1, |z-1|=r(0<r<1),$ then $$|\frac{\log (1-z)}{z}|\leq \dfrac{\log\frac1r}{1-r},$$ then show that $\lim\limits_{r\to0}\int_{\gamma_r}\frac{\log (1-z)}{z}dz=0$
2) definition $\int_{|z|=1}\frac{\log (1-z)}{z}dz=\lim\limits_{r\to0}\int_{\gamma_1}\frac{\log (1-z)}{z}dz $. Compute the integration
$$\int_{|z|=1}\frac{\log (1-z)}{z}dz$$
and then show that $\int_0^{2\pi}\log|1-e^{i\theta}|d\theta=0$
For 2), $\frac{\log{(1-z)}}{z}$ has no poles in the region bounded by $\gamma_1$. Therefore, by Cauchy's integral theorem,
$$\oint_{\gamma_1} dz \frac{\log{(1-z)}}{z} = 0 $$
From part 1), this implies that
$$\oint_{|z|=1} dz \frac{\log{(1-z)}}{z} = 0 $$
Let $z=e^{i \theta}$, $dz/z = i d\theta$, then
$$i \int_0^{2 \pi} d\theta \log{(1-e^{i \theta})} = 0$$
Now
$$\log{(1-e^{i \theta})} = \log{|1-e^{i \theta}|} +i \arg{(1-e^{i \theta})}$$
It turns out that
$$\arg{(1-e^{i \theta})} = \arctan{\frac{\sin{\theta}}{1-\cos{\theta}}} = \arctan{(\cot{\frac{\theta}{2}})} = \frac{\pi}{2} - \frac{\theta}{2}$$
$$\int_0^{2 \pi} d\theta \: \arg{(1-e^{i \theta})}= \int_0^{2 \pi} d\theta \left ( \frac{\pi}{2} - \frac{\theta}{2} \right ) = \frac{\pi}{2} 2 \pi - \frac{1}{4} (2 \pi)^2 = 0 $$
$$\therefore \int_0^{2 \pi} d\theta \log{|1-e^{i \theta}|} = 0$$