Yet another solution of an old problem

Consider a problem we already solved (in a weaker setting) in a previous post.
Let $x_1,\ldots, x_n$ be complex numbers, such that $$x_1^k+\cdots +x_n^k=0$$ for all $k\in\{1,\ldots,n\}$. Prove that all of the numbers are zeros.

Here we would investigate another approach. The idea is simple. First we construct a matrix $A$ whose characteristic polynomial has as roots $x_1,\ldots, x_n$. Then we use the fact that $\text{Tr}(A^k)=x_1^k+\cdots+x_n^k$ to successively eliminate all the coefficients, hence showing that $x_1,\ldots,x_n$ are roots of $x^n=0$.
Assume $x^n-a_1 x^{n-1}-\cdots-a_{n-1}x-a_n$ is a polynomial whose roots are precisely $x_1,\ldots,x_n$. Then the matrix
$$  A= \begin{pmatrix}
        a_1       & a_2 & a_3 & \dots & a_{n-1} & a_n \\
        1       & 0 & 0 & \dots  &0 & 0 \\
         0     & 1 & 0  & \dots &0 & 0 \\
         \vdots          &\vdots & \vdots &\vdots   &\ddots   &\vdots\\
        0      & 0 & 0 & \dots &1  & 0
    \end{pmatrix}
 $$ has this polynomial as characteristic.
Since $\text{Tr}(A)=0$ we have $a_1=0$. Now assume we have deduced that $a_1=\cdots=a_k=0$ for some $k\le n-1$.  It is easy to see by induction that $$A^{k+1}=\begin{pmatrix}
        a_{k+1}      & a_{k+2}  & a_{k+3} & \dots & \dots & a_{n}     & 0 & \dots & 0\\
        0    & a_{k+1} & a_{k+2}        & \dots  & \dots & a_{n-1}              & a_n &\dots       & 0\\   
    \vdots   &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots\\
       0       & 0 & 0        & \dots  &a_{k+1}              & \dots     & \dots & \dots & a_n  \\
1      & 0 & 0        & \dots  &\dots            &\dots &\dots &\dots &0  \\
0       & 1 & 0        & \dots  &\dots &\dots &\dots &\dots  &0\\

    \vdots   &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots&\vdots&\vdots\\

          0      & 0 & \dots & 1 & 0 & \dots & \dots &\dots & 0
    \end{pmatrix}$$
 (the matrix may not be perfectly written, but you got the feeling)
From here is evident that $\text {Tr}(A^{k+1})=(k+1) a_{k+1}$. Hence $(k+1)a_{k+1}=0$ thus $a_{k+1}$ is zero. Thus we obtain that all the coefficients are zero. Hence the roots.

Comment. Does this solution generalizes the problem into a more abstract setting?