For example, solve the system of equations
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=n, \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=n, \\
\cdots \cdots \\
x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}=n .
\end{array}\right.
$$
Consider the polynomial with roots $x_{1}, x_{2}, \cdots, x_{n}$
$$
\begin{aligned}
f(x) & =\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right) \\
& =x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} .
\end{aligned}
$$

Then
$$
\begin{array}{c}
f\left(x_{k}\right)=x_{k}^{n}+a_{n-1} x_{k}^{n-1}+\cdots+a_{1} x_{k}+a_{0}=0 \quad(k=1,2, \cdots, n) \\
\sum_{k=1}^{n} f\left(x_{k}\right)=\sum_{k=1}^{n}\left(x_{k}^{n}+a_{n-1} x_{k}^{n-1}+\cdots+a_{1} x_{k}+a_{0}\right) \\
=\sum_{k=1}^{n} x_{k}^{n}+a_{n-1} \sum_{k=1}^{n} x_{k}^{n-1}+\cdots+a_{1} \sum_{k=1}^{n} x_{k}+\sum_{k=1}^{n} a_{0}=0 . \\
n+a_{n-1} \cdot n+\cdots+a_{1} n+a_{0} n=0, \\
f(1)=1+a_{n-1}+\cdots+a_{1}+a_{0}=0 .
\end{array}
$$

From the system of equations, we know
Thus
This shows that $x=1$ is a root of $f(x)$. Without loss of generality, let $x_{n}=1$, from the original system of equations we get
$$
x_{1}^{k}+x_{2}^{k}+\cdots+x_{n-1}^{k}=n-1(k=1,2, \cdots, n-1) .
$$

Similarly, we can deduce that one of $x_{1}, \cdots, x_{n-1}$ is 1. Continuing this process, we must have
$$
x_{1}=x_{2}=\cdots=x_{n}=1 .
$$