If $\{x_i\}_{i=1}^n$ are the roots of $f(x)=a_nx^n + a_{n-1}x^{n-1} + \ldots +a_0$ then $\sum_{i=1}^nx_i^{n-1}$ is independent of $a_0$ I found an interesting conclusion when I did this simple question. Let
$$f(x)=(x^2-1)(x+2)=x^3+2x^2-x-2$$
and let $x_i$ for $i=1,2,3$ be the roots of $f(x)$. Find the sum $\sum\limits_{i=1}^3x_i^2$.
Obviously the answer is $1^2+(-1)^2+2^2=6$.
Now we change the constant term to $0$ s.t. $f(x)=x^3+2x^2-x$.
Then we find $\sum\limits_{i=1}^3x_i^2=0^2+(-1-\sqrt 2)^2+(-1+\sqrt 2)^2=6$.
It's interesting, for a simpler example $x^2-a^2=0$ we have two roots, $\forall a\in \mathbb{R}$, $x_1+x_2=0$ the sum will never change its value.
So I got a general conclusion:

If $a_0,a_1\cdots a_n\in \mathbb{R},~a_n\neq 0,~f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ and $x_i$ for $i=1,2\ldots n$ are the roots of $f(x)$ then whatever the value of $a_0$ is, the value of $\sum\limits_{i=1}^n x_i^{n-1}$ will never change! 

Maybe by changing the field $\mathbb{R}$ to $\mathbb{C}$ the result is also true? Who can prove it? Thanks!
 A: Let $\sigma_i(x_1,\ldots,x_n)$ be the $i$th order basic symmetric polynomial of $x_1,\ldots,x_n$.
Since:
$$
a_{n-i}=\sum_{1\leq k_1<\ldots<k_i\leq n} \prod_{j=1}^i -x_{k_j}=:\sigma_i(-x_1,\ldots,-x_n)
$$
And by the fundamental theorem of the symmetric polyinomials, there is a polynomial such that:
$$
\sum_{i=1}^n x^k_i=f(\sigma_1(-x_1,\ldots,-x_n),\ldots,\sigma_k(-x_1,\ldots,-x_n))=f(a_{n-1},\ldots,a_{n-k})
$$
For $k=n-1$ we have your question.
A: W.l.o.g. assume $a_n=1$. Then by Vieta's formulas we have $-a_{n-1}=x_1+x_2+\dotsc+x_n, a_{n-2}=x_1x_2+x_1x_3+\dotsc+x_{n-1}x_n, \dotsc, (-1)^na_0=x_1x_2\dotsc x_n$
These are called the elementary symmetric polynomials.
Now, you might observe that
$$x_1+x_2+\dotsc+x_n=-a_{n-1}$$
$$x_1^2+x_2^2+\dotsc+x_n^2=a_{n-1}^2-2a_{n-2}$$
Furthermore, one might obtain
$$x_1^3+x_2^3+\dotsc+x_n^3=-a_{n-1}^3+3a_{n-1}a_{n-2}-a_{n-3}$$
Now, one might suspect that these will be continued and indeed, they do.
This can be proved easily by induction and it's called the Newton Identities.
Now, it's clear that the such identity for $\sum_{i=1}^{n} x_i^{n-1}$ can't involve the $a_0$ term since it is of degree $n$ while the Newton Formulas are homogenous (in $x_1,x_2,\dotsc,x_n$).
So this sum can be expressed in terms of $a_{n-1},a_{n-2},\dotsc,a_1$ and is independent of $a_0$ as you suggested.
A: For a monic polynomial $p(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+\color{red}{a_0}$ take the Frobenius companion matrix
$$
A=\left[\matrix{0 & 0 & \ldots & 0 & \color{red}{-a_0}\\
1 & 0 & \ldots & 0 & -a_1\\
0 & 1 & \ldots & 0 & -a_2\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & \ldots & 1 & -a_{n-1}\\
}\right]=
\left[\matrix{0 & 0 & \ldots & 0 & \color{red}{*}\\
1 & 0 & \ldots & 0 & *\\
0 & 1 & \ldots & 0 & *\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & \ldots & 1 & *\\
}\right]
$$
where $\color{red}{*}$ are the elements that depend on $\color{red}{a_0}$ and $*$ are the elements that do not. We need to prove that $\text{tr}\,A^{n-1}$ does not depend on $\color{red}{a_0}$.
Proof: Let's see how the powers of $A$ depend on $\color{red}{a_0}$:
$$
A\cdot A=
\left[\matrix{0 & 0 & 0 & \ldots & 0 & \color{red}{*}\\
1 & 0 & 0 & \ldots & 0 & *\\
0 & 1 & 0 & \ldots & 0 & *\\
0 & 0 & 1 & \ldots & 0 & *\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \ldots & 1 & *\\
}\right]
\left[\matrix{0 & 0 & 0 & \ldots & 0 & \color{red}{*}\\
1 & 0 & 0 & \ldots & 0 & *\\
0 & 1 & 0 & \ldots & 0 & *\\
0 & 0 & 1 & \ldots & 0 & *\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \ldots & 1 & *\\
}\right]=
\left[\matrix{0 & 0 & 0 &\ldots & \color{red}{*} & \color{red}{*}\\
0 & 0 & 0 &\ldots & * & \color{red}{*}\\
1 & 0 & 0 & \ldots & * & *\\
0 & 1 & 0 & \ldots & * & *\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 &\ldots & * & *\\
}\right]
$$
$$
A^2\cdot A=
\left[\matrix{0 & 0 & 0 &\ldots & \color{red}{*} & \color{red}{*}\\
0 & 0 & 0 &\ldots & * & \color{red}{*}\\
1 & 0 & 0 & \ldots & * & *\\
0 & 1 & 0 & \ldots & * & *\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 &\ldots & * & *\\
}\right]
\left[\matrix{0 & 0 & 0 & \ldots & 0 & \color{red}{*}\\
1 & 0 & 0 & \ldots & 0 & *\\
0 & 1 & 0 & \ldots & 0 & *\\
0 & 0 & 1 & \ldots & 0 & *\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \ldots & 1 & *\\
}\right]=
\left[\matrix{0 & 0 & \ldots & \color{red}{*} & \color{red}{*} & \color{red}{*}\\
0 & 0 & \ldots & * & \color{red}{*} & \color{red}{*}\\
0 & 0 &  \ldots & * & * & \color{red}{*}\\
1 & 0 &  \ldots & * & * & *\\
\vdots & \vdots &  \ddots & \vdots & \vdots & \vdots\\
0 & 0 & \ldots & * & * & *\\
}\right]
$$
Post-multiplication by $A$ works very easy: the first $n-1$ columns of $A$ perform the left shift in each row of the first matrix, and only the last column mixes the elements up, however the red elements are easy to track - they appear in the same rows where they were before plus one more row where the first matrix has the identity in the first column.
Doing it sufficiently many times we get
$$
A^{n-1}=\left[\matrix{0 & \color{red}{*} & \color{red}{*} &\ldots & \color{red}{*} & \color{red}{*}\\
0 & * & \color{red}{*} &\ldots & \color{red}{*} & \color{red}{*}\\
0 & * & * & \ldots & \color{red}{*} & \color{red}{*}\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & * & * &\ldots & * & \color{red}{*}\\
1 & * & * &\ldots & * & *\\
}\right].
$$
Since the diagonal is free of the red elements, it contains no $\color{red}{a_0}$, so does the trace, which is equal to the sum of eigenvalues of $A^{n-1}$, that is $\sum_{k=1}^n x_k^{n-1}$.
