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?

Use induction:
for $n=2$, we have
 $$\big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_1}{z_2} \big)= 2 + \frac{z_2}{z_1} +  \frac{z_1}{z_2} \in\mathbb{R}  $$
since $ \dfrac{z_2}{z_1} = \exp\big( i \theta \big)$ for some $\theta\in\mathbb{R}$ while  $\dfrac{z_1}{z_2} = \exp\big( - i\theta \big)$ and $ \exp\big(  i\theta \big) +  \exp\big( - i\theta \big) = 2 \cos\theta$ is real.
Assume the result holds true for $k = n \geq 2$, that is, 
$$\big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_3}{z_2} \big)\ldots \big( 1 + \frac{z_n}{z_{n-1}} \big) \cdot \big( 1 + \frac{z_1}{z_n} \big)\in\mathbb{R}  ,   $$
then for $k = n+1$, we have 
\begin{align}
& \qquad \big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_3}{z_2} \big)\ldots \big( 1 + \frac{z_n}{z_{n-1}} \big) \cdot  \big( 1 + \frac{z_{n+1}}{z_n} \big)  \big( 1 + \frac{z_1}{z_{n+1}} \big) \\
& = \left[ \big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_3}{z_2} \big)\ldots \big( 1 + \frac{z_n}{z_{n-1}} \big) \cdot \big( 1 + \frac{z_1}{z_n} \big) \right]\times \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big)  \big( 1 + \dfrac{z_1}{z_{n+1}} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big)}
\end{align}
now it suffices to show $ \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big)  \big( 1 + \dfrac{z_1}{z_{n+1}} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big)}$ is a real number.
Clearly, $ \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big)  \big( 1 + \dfrac{z_1}{z_{n+1}} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big)} =   \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big)  \big( 1 + \dfrac{z_1}{z_{n+1}} \big) \big( 1 + \dfrac{z_n}{z_1} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big)  \big( 1 + \dfrac{z_n}{z_1} \big)}$ is real, since the denominator is real by step-1 and the nominator is real by some easy calculation.    Q.E.D
