Find image of $|z|\leq1$ For $$w=\prod_{k=1}^{n} \frac{z_k - z}{1-\bar{z_k}z}$$ where $0<|z_k|<1$
I know if I write $w=f_1 f_2...f_n$ for $f_k=\frac{z_k -z}{1-\bar{z_k}z}$ then $f_k$ maps $|z|\leq 1$ onto itself. thus f must map $|z|\leq 1$ to itself. I think $f$ will be also onto since $n=1$ is such a case. but I don't see how to proceed.
Use the argument principle. Since $f$ is holomorphic on a neighbourhood of the closed unit disk, and $\lvert f(z)\rvert = 1$ for $\lvert z\rvert = 1$, the number of times $f$ attains $w$ (counting multiplicity) in the open unit disk is given by
$$N(w) = \frac{1}{2\pi i} \int_{\lvert z\rvert = 1} \frac{f'(\zeta)}{f(\zeta) - w}\,d\zeta$$
for $w \in \mathbb{C}$ with $\lvert w\rvert \neq 1$. As a continuous integer-valued function $N$ is locally constant on its domain $\mathbb{C}\setminus \{w : \lvert w\rvert = 1\}$, so it is in particular constant on the open unit disk $\mathbb{D}$. But clearly $N(f(0)) > 0$, so $N(w) > 0$ for all $w \in \mathbb{D}$. Thus
$$\mathbb{D} \subset f(\mathbb{D})\,,$$
and since $\overline{\mathbb{D}}$ is compact we have
$$\overline{\mathbb{D}} \subset \overline{f(\mathbb{D})} = f(\overline{\mathbb{D}})\,.$$