Dynamical system $x_{n+1} = \frac{1}{2}(x_n - \frac{1}{x_n}) \ \ , \ \ n = 0, 1 , 2,...$ 
Consider the dynamical system
  $$
x_{n+1} = \frac{1}{2}(x_n - \frac{1}{x_n}) \ \ , \ \ n = 0, 1 , 2,...
$$

So by using the substitution $x_n = \cot(y_n)$, I have found:
$$
x_n = \cot(\cot^{-1} (x_0) \cdot 2^n )
$$
but the next part of the question says

Let us parametrize the initial condition $x_0$ by means of a unique angle $\theta \in (0, \pi)$ such that $x_0 = \cot(\theta)$. Show that, for every $p>1$, the choice
  $$
\theta = \theta_p := \frac{\pi}{2^p - 1}
$$
  yields a period-p solution.

So I know I have to show that $x_n = x_{n+p}$, that is:
$$
cot(\frac{\pi}{2^p - 1} \cdot 2^n) = cot(\frac{\pi}{2^p - 1} \cdot 2^{n+p})
$$
and so, using periodicity of $\cot$, that
$$
\frac{\pi}{2^p - 1} \cdot 2^n + k \pi= \frac{\pi}{2^p - 1} \cdot 2^{n+p}
$$
for some $k\in\mathbb{Z}$. Inductively, we have for $p = 1$ that $\theta_1 = \pi$ hence
$$
x_{n+1} = \cot (\pi \cdot 2^{n+1}) = \cot(\pi \cdot 2^n + \pi \cdot 2^n) = \cot(\pi \cdot 2^n) = x_n
$$
and so we have a period-1 solution. How do I show the inductive step for this proof?

For $x_0=\cot(\theta)$ we have that 
$x_1=\frac{1}{2}\Big[\frac{\cos(\theta)}{\sin(\theta)}-\frac{\sin(\theta)}{\cos(\theta)}\Big]= \cot(2\theta)$
and inductively
$x_{n}=\cot(2^n\theta)$
Thus when $\theta=\frac{\pi}{2^p-1}$ we obtain
$x_{n+p}=\cot\big(\frac{2^{n+p}\pi}{2^p-1}\big)=\cot\Big(\frac{2^n(2^{p}-1)\pi}{2^p-1}+\frac{2^{n}\pi}{2^p-1}\Big)=\cot\Big(2^n\pi+\frac{2^{n}\pi}{2^p-1}\Big)=\cot\Big(\frac{2^{n}\pi}{2^p-1}\Big)=x_n$
since $2^n\pi$ ia an integer multiple of $\pi$.
