Denote $$A = (0, +\infty) × (0, π)^{n - 2} × (-π, π),\\ B = \{(x_1, \cdots, x_n) \in \mathbb{R}^n \mid x_{n - 1} \leqslant 0,\ x_n = 0\}.$$ It will be proved that $\mathrm{im}({\mit Ψ}) = \mathbb{R}^n\setminus B$.
First, consider any $(x_1, \cdots, x_n) \in B$. Suppose there exists $(r, θ_1, \cdots, θ_{n - 1}) \in A$ such that $$(x_1, \cdots, x_n) = {\mit Ψ}(r, θ_1, \cdots, θ_{n - 1}).$$
Because $x_n = r \sin θ_1 \cdots \sin θ_{n - 2} \sin θ_{n - 1} = 0$ and$$ r > 0,\ θ_1, \cdots, θ_{n - 2} \in (0, π) \Longrightarrow r \sin θ_1 \cdots \sin θ_{n - 2} > 0, $$ then $\sin θ_{n - 1} = 0$. Note that $θ_{n - 1} \in (-π, π)$, thus $θ_{n - 1} = 0$. Therefore,$$ 0 \geqslant x_{n - 1} = r \sin θ_1 \cdots \sin θ_{n - 2} \cos θ_{n - 1} = r \sin θ_1 \cdots \sin θ_{n - 2} > 0, $$ a contradiction. Thus $B \cap \mathrm{im}({\mit Ψ}) = \varnothing$.
Now consider any $(x_1, \cdots, x_n) \not\in B$. Take $r = \sqrt{\sum\limits_{k = 1}^n x_k^2}$ and $θ_1 = \arccos \dfrac{x_1}{r} \in (0, π)$.
Suppose $θ_j$ has been defined for $1 \leqslant j \leqslant k - 1$ as$$ θ_j = \arccos \frac{x_j}{r \sin θ_1 \cdots \sin θ_{j - 1}}, $$where $k \leqslant n - 2$. Because$$ \sin^2 θ_j = 1 - \frac{x_j^2}{r^2 \sin^2 θ_1 \cdots \sin^2 θ_{j - 1}}\\ \Longrightarrow r^2 \sin^2 θ_1 \cdots \sin^2 θ_{j - 1} - r^2 \sin^2 θ_1 \cdots \sin^2 θ_j = x_j^2, \quad 1 \leqslant j \leqslant k - 1 $$ then\begin{align*} \sum_{j = 1}^{k - 1} x_j^2 &= \sum_{j = 1}^{k - 1} (r^2 \sin^2 θ_1 \cdots \sin^2 θ_{j - 1} - r^2 \sin^2 θ_1 \cdots \sin^2 θ_j)\\ &= r^2 - r^2 \sin^2 θ_1 \cdots \sin^2 θ_{k - 1}, \tag{1} \end{align*} which implies$$ r^2 \sin^2 θ_1 \cdots \sin^2 θ_{k - 1} = r^2 - \sum_{j = 1}^{k - 1} x_j^2 = \sum_{j = k}^n x_j^2 > x_k^2, \tag{2} $$ where the last inequality uses the fact that$$ (x_1, \cdots, x_n) \not\in B \Longrightarrow x_{n - 1} \neq 0 \text{ or } x_n \neq 0. $$ Now take$$ θ_k = \arccos \frac{x_k}{r \sin θ_1 \cdots \sin θ_{k - 1}} \in (0, π). $$
At last, suppose $θ_1, \cdots, θ_{n - 2}$ have been defined as$$ θ_k = \arccos \frac{x_k}{r \sin θ_1 \cdots \sin θ_{k - 1}}. \quad 1 \leqslant k \leqslant n - 2 $$ Case 1: $x_n = 0$. Take $θ_{n - 1} = 0 \in (-π, π)$, then$$ r \sin θ_1 \cdots \sin θ_{n - 1} = 0 = x_n. $$ Analogous to (1), there is$$ \sum_{k = 1}^{n - 2} x_k^2 = r^2 - r^2 \sin^2 θ_1 \cdots \sin^2 θ_{n - 2}. $$ Note that $(x_1, \cdots, x_n) \not\in B$, then $x_{n - 1} > 0$. Since $x_n = 0$, then$$ r \sin θ_1 \cdots \sin θ_{n - 2} = \sqrt{r^2 - \sum_{k = 1}^{n - 2} x_k^2} = \sqrt{x_{n - 1}^2 + x_n^2} = x_{n - 1}. $$ Therefore$$ {\mit Ψ}(r, θ_1, \cdots, θ_{n - 1}) = (x_1, \cdots, x_n) \Longrightarrow (x_1, \cdots, x_n) \in \mathrm{im}({\mit Ψ}). $$
Case 2: $x_n > 0$. Analogous to (2), there is$$ r^2 \sin^2 θ_1 \cdots \sin^2 θ_{n - 2} = x_{n - 1}^2 + x_n^2 > x_{n - 1}^2. $$ Take$$ θ_{n - 1} = \arccos \frac{x_{n - 1}}{r \sin θ_1 \cdots \sin θ_{n - 2}} \in (0, π). $$ Again analogous to (1), there is$$ \sum_{k = 1}^{n - 1} x_k^2 = r^2 - r^2 \sin^2 θ_1 \cdots \sin^2 θ_{n - 1}. $$ Note that $x_n > 0$, thus$$ r \sin θ_1 \cdots \sin θ_{n - 1} =\sqrt{r^2 - \sum_{k = 1}^{n - 1} x_k^2} = x_n. $$ Therefore$$ {\mit Ψ}(r, θ_1, \cdots, θ_{n - 1}) = (x_1, \cdots, x_n) \Longrightarrow (x_1, \cdots, x_n) \in \mathrm{im}({\mit Ψ}). $$
Case 3: $x_n < 0$. The proof for this case is the same as that for Case 2 except for changing $$θ_{n - 1} = \arccos \frac{x_{n - 1}}{r \sin θ_1 \cdots \sin θ_{n - 2}} \in (0, π)$$ to$$ θ_{n - 1} = -\arccos \frac{x_{n - 1}}{r \sin θ_1 \cdots \sin θ_{n - 2}} \in (-π, 0). $$
Therefore, $\mathrm{im}({\mit Ψ}) = \mathbb{R}^n \setminus B$.
