Let $x=(x_1, \cdots, x_n)\in \mathbb{R}^n$ be Cartesian coordinates and $(r,\theta)=(r,\theta_1, \cdots , \theta_{n-1})$, where $r\in (0,\infty), \theta_1,\dots, \theta_{n-2}\in (0,\pi)$ and $\theta_{n-1}\in (-\pi,\pi)$ be polar coordinates. The coordinate transform is given by $$\Psi:(r,\theta_1, \cdots, θ_{n - 1})\mapsto x=\begin{cases} x_1=r\cos{\theta_1}, \\ x_2=r\sin{\theta_1}\cos{\theta_2}, \\ x_3=r\sin{\theta_1}\sin{\theta_2}\cos{\theta_3}, \\ \vdots \\ x_{n-1}=r\sin{\theta_1}\sin{\theta_2}\dots \sin{\theta_{n-2}}\cos{\theta_{n-1}}, \\ x_n=r\sin{\theta_1}\sin{\theta_2}\dots \sin{\theta_{n-2}}\sin{\theta_{n-1}}. \end{cases}$$
I am trying to find the range of this function $\Psi$. I think it is $\mathbb{R}^n \backslash \{x:x_n=0,x_{n-1}\le 0\}$, but I cannot think of a way to show this explicitly from the definition of $\Psi$. I would greatly appreciate any help.
Show that for all $(r, \theta_1, \ldots, \theta_{n-1}) \in (0, \infty) \times (0, \pi)^{n-2} \times (-\pi, \pi)$, $\Psi(r, \theta) \in \mathbb{R}^n \setminus \{x:x_n =0, x_{n-1} \leq 0\}$.
This merely follows from $x_n=0$ requiring that $\theta_{n-1}=0$.
Then show that for all $x \in \Psi(r, \theta) \in \mathbb{R}^n \setminus \{x:x_n =0, x_{n-1} \leq 0\}$, there exists $(r, \theta_1, \ldots, \theta_{n-1}) \in (0, \infty) \times (0, \pi)^{n-2} \times (-\pi, \pi)$ such that $x = \Psi(r, \theta)$.
The trick is to this is to define $r, \theta$ in terms of $x$. The following should work for $r$: $$r = \sqrt{\sum_{i=1}^nx_i^2}.$$ $\theta$ can be defined recursively.