[proofplan] We prove by induction on $n$ that the [Jacobian](/page/Derivative) determinant $J_n$ of the transformation from $n$-dimensional spherical coordinates to Cartesian coordinates satisfies $J_n = r^{n-1}\prod_{j=1}^{n-2}\sin^{n-j-1}(\theta_j)$. The base cases $n = 2$ and $n = 3$ are computed directly. For the inductive step, we expand the $n \times n$ determinant by cofactors along the first row, factor common terms from the resulting $(n-1) \times (n-1)$ minors, recognise each minor as $J_{n-1}$, and simplify using the Pythagorean identity $\cos^2\theta_1 + \sin^2\theta_1 = 1$. [/proofplan] [step:Set up notation and write the Jacobian matrix] For brevity, write $c_j := \cos\theta_j$ and $s_j := \sin\theta_j$. The transformation from [spherical coordinates](/page/Spherical%20Integrals%20and%20Radial%20Functions) $(r, \theta_1, \dots, \theta_{n-1})$ to Cartesian coordinates $(x_1, \dots, x_n)$ is given by: \begin{align*} x_1 &= r\,c_1, \\ x_2 &= r\,s_1\,c_2, \\ &\;\;\vdots \\ x_{n-1} &= r\left(\prod_{j=1}^{n-2}s_j\right)c_{n-1}, \\ x_n &= r\prod_{j=1}^{n-1}s_j. \end{align*} The Jacobian [determinant](/page/Derivative) is: \begin{align*} J_n(r, \theta_1, \dots, \theta_{n-1}) := \det\left(\frac{\partial x_i}{\partial r} \;\bigg|\; \frac{\partial x_i}{\partial \theta_j}\right)_{\substack{1 \le i \le n \\ 1 \le j \le n-1}}. \end{align*} [/step] [step:Expand the determinant by cofactors along the first row] The first row of the Jacobian matrix is $(c_1, -rs_1, 0, \dots, 0)$, since $\partial x_1/\partial r = c_1$, $\partial x_1/\partial \theta_1 = -rs_1$, and $\partial x_1/\partial \theta_j = 0$ for $j > 1$. Expanding by cofactors along the first row: \begin{align*} J_n = c_1 \det A + r\,s_1 \det B, \end{align*} where $A$ is the $(n-1) \times (n-1)$ minor obtained by deleting row 1 and column 1, and $B$ is the $(n-1) \times (n-1)$ minor obtained by deleting row 1 and column 2 (with the appropriate sign from cofactor expansion absorbed). [guided] The key observation is that the first row has only two non-zero entries: $c_1$ in column 1 and $-rs_1$ in column 2. All remaining entries in row 1 are zero because $x_1 = r\,c_1$ depends only on $r$ and $\theta_1$, not on $\theta_2, \dots, \theta_{n-1}$. Cofactor expansion along row 1 gives: \begin{align*} J_n = c_1 \cdot (-1)^{1+1} \det A + (-rs_1) \cdot (-1)^{1+2} \det B = c_1 \det A + rs_1 \det B, \end{align*} where $A$ is the $(n-1) \times (n-1)$ submatrix obtained by deleting row 1 and column 1, and $B$ is the submatrix obtained by deleting row 1 and column 2. The sign on the second term simplifies: $(-rs_1)(-1) = rs_1$. [/guided] [/step] [step:Factor common terms from $\det A$ and $\det B$ to recognise $J_{n-1}$] In the minor $A$, every entry in column 1 contains a factor of $rc_1$, and every entry in columns $2, \dots, n-1$ contains a factor of $s_1$. Factoring these out: \begin{align*} \det A = rc_1\,s_1^{n-2} \det A', \end{align*} where $A'$ is the $(n-1) \times (n-1)$ matrix with the same structure as the Jacobian matrix for $n-1$ dimensions with variables $(r, \theta_2, \dots, \theta_{n-1})$. That is, $\det A' = J_{n-1}(r, \theta_2, \dots, \theta_{n-1})$. Similarly, in the minor $B$, every column contains a factor of $s_1$. Factoring: \begin{align*} \det B = s_1^{n-1} \det B', \end{align*} where $\det B' = J_{n-1}(r, \theta_2, \dots, \theta_{n-1})$ by the same structural recognition. [guided] Consider the minor $A$. Its rows correspond to $x_2, \dots, x_n$ and its columns correspond to $\theta_1, \dots, \theta_{n-1}$. Each $x_i$ for $i \ge 2$ has the form $x_i = r \cdot s_1 \cdot (\text{product of } c_j, s_j \text{ for } j \ge 2)$. The derivative $\partial x_i / \partial r$ always carries the factor $c_1 \cdot (\text{angular part})$ from differentiating the original formula, while $\partial x_i / \partial \theta_j$ for $j \ge 2$ carries a factor $s_1$. More precisely, examining column 1 of $A$ (derivatives with respect to $r$): every entry has a factor $c_1$ since $\partial x_i/\partial r$ evaluated at $r$ gives the angular part with $c_1$ from the original Jacobian structure. Columns $2, \dots, n-1$ of $A$ (derivatives with respect to $\theta_2, \dots, \theta_{n-1}$): every entry has a factor $s_1$ since each $x_i$ for $i \ge 2$ contains $s_1$ as a factor, which persists under differentiation with respect to $\theta_j$ for $j \ge 2$. Factoring $rc_1$ from column 1 and $s_1$ from each of the remaining $n-2$ columns: \begin{align*} \det A = rc_1 \cdot s_1^{n-2} \cdot J_{n-1}(r, \theta_2, \dots, \theta_{n-1}), \end{align*} where $J_{n-1}$ is the Jacobian determinant for $(n-1)$-dimensional spherical coordinates. The remaining matrix after factoring has the same structure as the Jacobian for the transformation $(r, \theta_2, \dots, \theta_{n-1}) \mapsto (x_2', \dots, x_n')$ where $x_i' = x_i / (s_1)$. For $\det B$: every column of $B$ contains a factor of $s_1$ (column 1 because each $x_i$ for $i \ge 2$ contains $s_1$ as a factor of $\partial x_i / \partial r$, and columns $2, \dots, n-1$ for the same reason as above). Factoring: \begin{align*} \det B = s_1^{n-1} \cdot J_{n-1}(r, \theta_2, \dots, \theta_{n-1}). \end{align*} [/guided] [/step] [step:Combine using the Pythagorean identity to obtain the recurrence] Substituting the factored expressions: \begin{align*} J_n &= c_1 \cdot rc_1\,s_1^{n-2}\,J_{n-1} + rs_1 \cdot s_1^{n-1}\,J_{n-1} \\ &= rc_1^2\,s_1^{n-2}\,J_{n-1} + rs_1^n\,J_{n-1} \\ &= rs_1^{n-2}\,J_{n-1}\left(c_1^2 + s_1^2\right) \\ &= rs_1^{n-2}\,J_{n-1}, \end{align*} where we used the Pythagorean identity $c_1^2 + s_1^2 = \cos^2\theta_1 + \sin^2\theta_1 = 1$. [/step] [step:Verify the base cases and complete the induction] **Base case $n = 2$:** The transformation is $x_1 = r\cos\theta_1$, $x_2 = r\sin\theta_1$. The Jacobian matrix is: \begin{align*} \begin{pmatrix} \cos\theta_1 & -r\sin\theta_1 \\ \sin\theta_1 & r\cos\theta_1 \end{pmatrix}, \end{align*} with determinant $J_2 = r(\cos^2\theta_1 + \sin^2\theta_1) = r$. This matches $r^{2-1} \cdot (\text{empty product}) = r$. **Base case $n = 3$:** Direct computation gives $J_3 = r^2 \sin\theta_1$, matching $r^{3-1}\sin^{3-2}\theta_1 = r^2\sin\theta_1$. **Inductive step:** Assume $J_{n-1}(r, \theta_2, \dots, \theta_{n-1}) = r^{n-2}\prod_{j=2}^{n-2}\sin^{n-j-1}(\theta_j)$. By the recurrence: \begin{align*} J_n &= r\,\sin^{n-2}(\theta_1)\,J_{n-1}(r, \theta_2, \dots, \theta_{n-1}) \\ &= r \cdot \sin^{n-2}(\theta_1) \cdot r^{n-2}\prod_{j=2}^{n-2}\sin^{n-j-1}(\theta_j) \\ &= r^{n-1}\sin^{n-2}(\theta_1)\prod_{j=2}^{n-2}\sin^{n-j-1}(\theta_j) \\ &= r^{n-1}\prod_{j=1}^{n-2}\sin^{n-j-1}(\theta_j). \end{align*} This completes the induction. [guided] Let us verify that the recurrence $J_n = r\,s_1^{n-2}\,J_{n-1}$ together with the base cases yields the claimed closed-form formula. **Base case $n = 2$:** The Jacobian matrix is $2 \times 2$: \begin{align*} \begin{pmatrix} \cos\theta_1 & -r\sin\theta_1 \\ \sin\theta_1 & r\cos\theta_1 \end{pmatrix}. \end{align*} The determinant is $r\cos^2\theta_1 + r\sin^2\theta_1 = r$. The claimed formula gives $r^{2-1} \cdot \prod_{j=1}^{0}(\cdots) = r \cdot 1 = r$, since the empty product equals $1$. The base case holds. **Base case $n = 3$:** Using the recurrence, $J_3 = r\,\sin^1(\theta_1)\,J_2 = r\sin\theta_1 \cdot r = r^2\sin\theta_1$. The claimed formula gives $r^2\sin^{3-1-1}(\theta_1) = r^2\sin\theta_1$. The base case holds. **Inductive step:** Suppose the formula holds for $n-1$, i.e., $J_{n-1}(r, \theta_2, \dots, \theta_{n-1}) = r^{n-2}\prod_{j=2}^{n-2}\sin^{n-j-1}(\theta_j)$. Note that the product starts at $j = 2$ because the $(n-1)$-dimensional spherical coordinate system uses angles $\theta_2, \dots, \theta_{n-1}$, and the exponent of $\sin\theta_j$ in $J_{n-1}$ is $(n-1) - (j-1) - 1 = n - j - 1$. Applying the recurrence: \begin{align*} J_n &= r \cdot \sin^{n-2}(\theta_1) \cdot r^{n-2}\prod_{j=2}^{n-2}\sin^{n-j-1}(\theta_j) = r^{n-1}\prod_{j=1}^{n-2}\sin^{n-j-1}(\theta_j), \end{align*} where we merged $\sin^{n-2}(\theta_1) = \sin^{n-1-1}(\theta_1)$ into the product as the $j = 1$ term. This completes the induction and proves the formula for all $n \ge 2$. [/guided] [/step]