Prove that the complex expression is real Let $|z_1|=\dots=|z_n|=1$ on the complex plane.
Prove that:
$$
\left(1+\frac{z_2}{z_1}\right)
\left(1+\frac{z_3}{z_2}\right)
\dots
\left(1+\frac{z_n}{z_{n-1}}\right)
\left(1+\frac{z_1}{z_n}\right)
\in\mathbb{R}
$$
I have tried induction and writing every "subexpression" as $(1+e^{i(\theta_n-\theta_{n-1})})$.
Any ideas?

Writing $z_k = e^{i\theta_k}$, $k = 1,2,\ldots, n$, we write the above expression as 
\begin{align}&(1 + e^{i(\theta_2 - \theta_1)})(1 + e^{i(\theta_3 - \theta_2)})\cdots (1 + e^{i(\theta_n - \theta_{n-1})})(1 + e^{i(\theta_1 - \theta_n)})\\
&= e^{-i(\theta_2 - \theta_1)/2}e^{-i(\theta_3 - \theta_2)/2}\cdots e^{-i(\theta_n - \theta_{n-1})/2}(e^{-i(\theta_2- \theta_1)/2} + e^{i(\theta_2 - \theta_1)/2})\cdots (e^{-i(\theta_1 - \theta_n)/2} + e^{i(\theta_1 - \theta_n)/2})\\
&= e^{-i[(\theta_2 - \theta_1) + (\theta_3 - \theta_2) + \cdots + (\theta_1 - \theta_n)]/2}\cdot 2^n\cos((\theta_2 - \theta_1)/2)\cos((\theta_3 - \theta_2)/2)\cdots \cos((\theta_1 - \theta_n)/2)\\
&= 2^n\cos((\theta_2 - \theta_1)/2)\cos((\theta_3 - \theta_2)/2)\cdots \cos((\theta_1 - \theta_n)/2).
\end{align}
The last expression is a real number.
