We know $$\begin{align}Log(1-z)&=-\sum_{n=1}^{\infty}\frac{z^n}{n}=-(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)\\ \Rightarrow [Log(1-z)]^2&=[-(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)]^2 \\&=(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)^2 \\ &=(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots)\\ &=0+(0)z+(1)z^2+(1)z^3+\frac{11}{12}z^4+\frac{5}{6}z^5+\cdots\end{align}$$
Series is valid at $|z|<1$.