system $$\omega^*_i
        -
                                a_{i,\pi(i)}' \,
        \left( \frac{x_i}{x_{\pi(i)}} - \frac{x_{\pi(i)}}{x_i} \right)
        = 0
        \quad \text{ for } i = 1,\dots,n
        \label{equ:tree-reduced}$$ for some complex constants $\omega^*_1,\dots,\omega^*_n$.

Here, the equivalence means the two systems have the same solution set in ${\mathbb{C}}^*$.

For $N=2$, the system  contains only one equation $$\omega_i - a_{1,0}' (x_1 x_0^{-1} - x_1^{-1} x_0) = 0$$ where $x_0 = 1$. The statement is obviously true in this case.

Now consider a tree $T_N$ consisting of $N = n+1$ nodes and assume the statement is true for any tree of smaller sizes. Fixing any leaf vertex in the tree, without loss, we can re-index the vertices so that this leaf vertex has index $n$ and its unique parent vertex is $n-1$. In this arrangement, the $n$-th (last) equation in  is $$\omega_n - a_{n,n-1}' (x_n x_{n-1}^{-1} - x_{n-1} x_n^{-1}) = 0,
       \label{equ:leaf-x}$$ while the $(n-1)$-th equation is $$\omega_{n-1} -
       a_{n-1,n}' (x_{n-1} x_n^{-1} - x_n x_{n-1}^{-1}) -
       \sum_{j \in \mathcal{N}(n-1) \setminus \{n\}}
       a_{n-1,j}' (x_{n-1} x_j^{-1} - x_j x_{n-1}^{-1})
       = 0.
       \label{equ:leaf-parent}$$ Then adding $a_{n-1,n}'/a_{n,n-1}'$ times  to  produces $$\left(\omega_{n-1} + \frac{a_{n-1,n}'}{a_{n,n-1}'} \omega_n \right)
       -
       \sum_{j \in \mathcal{N}(n-1) \setminus \{n\}}
       a_{n-1,j}' \left(\frac{x_{n-1}}{x_j} - \frac{x_j}{x_{n-1}} \right)
       = 0.$$ With this transformation, the first $n-1$ equations 