Note: I asked this on MSE as well, but I started thinking MO might be a better place to post the question.
$\textbf{Notation:}$
$\bullet$ For $N$ vectors $\textbf{u}_1, \ldots, \textbf{u}_N \in S^{n-1} \subset \mathbb{R}^n$, let $\alpha_{ij}$ ($i < j$) denote the angle between $\textbf{u}_i$ and $\textbf{u}_j$
$\bullet$ Let $\omega_n$ be the surface area (ie. $(n-1)$-dimensional volume) of the sphere $S^{n-1}$.
$\bullet$ Let $\sigma$ be the surface measure on $S^{n-1}$.
$\bullet$ Let $I( \cdots)$ be the indicator function.
$\textbf{Setup:}$
I'm reading an article in which the author is considering the integral \begin{align} \frac{1}{\omega_n^N} \cdot \int_{S^{n-1}} \cdots \int_{S^{n-1}} I(d_{ij} < \alpha_{ij} \leqslant d_{ij}' : 1 \leqslant i < j \leqslant N ) \hspace{0.1cm} d \sigma(\textbf{u}_1) \cdots d \sigma(\textbf{u}_N). \end{align} By using the spherical symmetry of the integrand, he then proceeds by rewriting (1) as \begin{align} \omega_n^{N-1} \prod_{k = 1}^{N-1} \omega_{n-k} \int_{\phi_{12}=0}^\pi \cdots \int_{\phi_{(N-1)N}=0}^\pi I(d_{ij} < \alpha_{ij} \leqslant d_{ij}' : 1 \leqslant i < j \leqslant N) \\ \nonumber \times \prod_{1 \leqslant i < j \leqslant N} \sin^{n-i-1} \phi_{ij} \hspace{0.1cm} d\phi_{(N-1)N} \cdots d \phi_{13} d \phi_{12} \end{align} ($\tfrac{1}{2}N(N-1)$ integrals in total), where now he only considers unit vectors of the form \begin{align*} \textbf{u}_1 &= (1, 0, \ldots, 0) \\ \textbf{u}_2 &= (\cos \phi_{12}, \sin \phi_{12}, 0, \ldots, 0) \\ \textbf{u}_3 &= (\cos \phi_{13}, \sin \phi_{13} \cos \phi_{23}, \sin \phi_{13} \sin \phi_{23}, 0 \ldots, 0) \\ &\hspace{0.2cm}\vdots \\ \textbf{u}_N &= (\cos \phi_{1N}, \sin \phi_{1N} \cos \phi_{2N}, \ldots, \sin \phi_{1N} \cdots \sin \phi_{(N-1)N}, 0, \ldots, 0). \end{align*} $\textbf{My problem:}$
The idea behind the transformation of (1) into (2) is that, by restricting the unit vector $\textbf{u}_1$ to a $1$-dimensional subspace, one has to compensate by adding a factor of $\omega_n$ in front of the integral; by restricting the unit vector $\textbf{u}_2$ to a $2$-dimensional subspace, one has to compensate by adding a factor of $\omega_{n-1}$ in front of the integral; and generally, by restricting $\textbf{u}_t$ to a $t$-dimensional subspace, one has to compensate by adding a factor of $\omega_{n+1-t}$ in front of the integral. This makes sense intuitively, but I haven't been able to actually prove it.
Any thoughts on this are very welcome!