Simplify product of two product operators

Question

Given the relation $$ \prod_{j=2}^{N-1}\left( \prod_{k=1}^{j-1} \sin^2\theta_k\right), $$ I want to show that the power of $\sin^2\theta_m$ is $\sum_{j=m+1}^{N-1}1=N-m-1$, knowing that $\theta_m$ appears once for each $j>m$. Hence, $$ \prod_{j=2}^{N-1}\left( \prod_{k=1}^{j-1} \sin^2\theta_k\right) = \prod_{j=1}^{N-2}\sin^{2(N-j-1)}\theta_j. $$ This result is part of the determinant of the metric tensor for the n-sphere. The article I'm reading does this simplification but I have no idea how it comes to that result. If anyone has some tips it would be of great help. Thanks in advance.

Answer 1

It is exactly as you said already. It does not really need a proof. Sometimes you look at something, and just do not see it. But you are right.

If you wish to make a more detailed proof than you already have, here. For each $j$ define $\alpha(j,k) = 1$ if $k<j$ and $\alpha(j,k)=0$ if $j\geq k$. Then you can write $$ \prod_{k=1}^{j-1} \sin^2(\theta_k)\, =\, \prod_{k=1}^{N-2} (\sin^2(\theta_k))^{\alpha(j,k)} $$ Then $$ \prod_{j=2}^{N-1} \left(\prod_{k=1}^{j-1} \sin^2(\theta_k)\right) $$ can be written as $$ \prod_{k=1}^{N-2} (\sin^2(\theta_k))^{\sum_{j=2}^{N-1} \alpha(j,k)} $$ and $\sum_{j=2}^{N-1} \alpha(j,k) = N-k-1$ as you already know (because it is $1$ only if $j>k$ and there are exactly $N-k-1$ choices of $j$ such that $k<j\leq N-1$).