Boundary behaviour of finite Blaschke products on the unit circle It is well known that a finite Blaschke product $$B(z)=\prod_{k=1}^n\left(\dfrac{z-a_k}{1-\overline{a_k}z}\right),\,\,\,\,\,\,\,\forall a_k\in\Bbb{D}$$ has exactly $n$ distinct per-images of any $\lambda\in\partial\mathbb{D}.$
Let us take them $(x_k)_{k=1}^n\in\mathbb{D},$ where $\arg x_1\lt\arg x_2\lt\cdots\lt\arg x_n\lt2\pi+\arg x_1.$
How can I prove that each arc $[x_k, x_{k+1})$ mapped bijectively on to the unit circle?  
I thought we can use to fact that the derivative of $B$ never vanishes on the unit circle. $$|B^{'}(z)|=\sum_{k=1}^n\dfrac{1-|a_k|^2}{|z-a_k|^2}\gt 0,\,\,\,\,\,\,\,\forall z\in\partial\Bbb{D}.$$ But I could not figure out a relationship between these properties.
Can anyone help me to prove this?
At least a hint regarding this?
 A: The abstract argument by Daniel is more elegant. If you want to do it by hand, however, here is a suggestion:
Without taking absolute values we have:
$$ \frac{B'(z)}{B(z)} =
\sum_{k=1}^n \left( \frac{1}{z-a_k}
 + \frac{\overline{a}_k}{1-\overline{a}_k z} \right) =
\sum_{k=1}^n 
  \frac{1-|a_k|^2}{(z-a_k)(1-\overline{a}_k z)} 
$$
The path $z=e^{it}$ then maps to
$$ \frac{B'(z)}{B(z)}    =  e^{-it} 
\sum_{k=1}^n   \frac{1-|a_k|^2}{|e^{it}-a_k|^2}  $$
or if we pose $B(t)=e^{i\phi(t)}$:
$$\phi'(t) = -i \frac{1}{B}\frac{dB}{dt}=z \frac{B'(z)}{B(z)}    =  
\sum_{k=1}^n   \frac{1-|a_k|^2}{|e^{it}-a_k|^2} >0 $$
The argument of $B(z(t))$ is thus monotonously increasing with $t$. The fact that it winds around $n$ times may be seen from calculating
$$ \int_0^{2\pi} \phi'(t)\frac{dt}{2\pi} = \int_0^{2\pi} \frac{1}{B}\frac{dB}{dt} \frac{dt}{2\pi i}= \oint \frac{B'(z)}{B(z)}\frac{dz}{2\pi i} = \oint \sum_{k=1}^n \left( \frac{1}{z-a_k}
 + \frac{\overline{a}_k}{1-\overline{a}_k z} \right) \frac{dz}{2\pi i}  = n $$
using the first formula and residue calculus (there are $n$ poles inside the contour). 
A simpler argument is that it must be an integer multiple of $2\pi$ and depend continuously upon the $a_k$'s. So now let all $a_k\rightarrow 0$ and you end up with $B(z)=z^n$ for which the result is obvious.
