To calculate the derivative of $$f(z)=\prod_{m=1}^n\frac{z-a_m}{1-\overline{a_m}z}$$ I have used $\frac{d}{dz}\prod_{m=1}^nf_m(z)=(\prod_{m=1}^nf_m(z))(\sum_{m=1}^n\frac{f_m'(z)}{f_m(z)})=f(z)(\sum_{m=1}^n\frac{1-|a_m|^2}{(1-\overline{a_m}z)(z-a_m)})$ where $f_m$ is one element in the product, but I do not know how to further simplify. In essence, I want to show that $f'(z)\neq0$ everywhere inside the unit disk, i.e. $|f'(z)|>0$.
I have seen here that $|f^{'}(z)|=\sum_{k=1}^n\dfrac{1-|a_k|^2}{|z-a_k|^2}$ and would like to get to this form.