Is there any way to prove the following conjecture regarding the Generalization of Quadratic Reciprocity Law. The statement being,
$$ \left(\dfrac{a_1}{a_2}\right)\left(\dfrac{a_2}{a_3}\right) \ldots\left(\dfrac{a_n}{a_1}\right)=(-1)^{\left(\frac{a_1-1}{2}\right) \left(\frac{a_2-1}{2}\right) \ldots \left(\frac{a_n-1}{2}\right)}$$
where the symbols are Jacobi Symbols and $\operatorname{gcd}(a_i,a_j)=1$ for $i\neq j$ and $a_i>1$ forall $i$ satisfying $1 \leq i \leq n$.
One of my friend told me that the argument for $n=3$ and when all $a_i$ are distinct primes, is a modification of the Eisenstein's argument for proving the Quadratic Reciprocity Law itself. Instead of considering a planer rectangle as in Eisenstein's Proof of Quadratic Reciprocity Law if we consider a cuboid and count the number of points $(x,y,z)$ (where $x$,$y$ and $z$ are all integers) that is within this cuboid and not on any outer surface, then the proof would be done. And when the result is proved for primes it can be proved in general also.
However the problem is that in this scheme we can't visualize an $n$-dimensional rectangle and thus can't forward his argument to $n$ dimensions, though we think that it can be generalized. Is there any way to help visualizing it or is there any way to prove the result in a simpler way?
If $a_1=5$, $a_2=3$, $a_3=7$ then all the Legendre symbols are $-1$, so their product is $-1$, but the right side is $(-1)^{2\times1\times3}=1$.