Example 6 Solve the system of equations $\left\{\begin{array}{l}x_{1}+x_{2}+\cdots+x_{n}=n \\ x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=n \\ \cdots \cdots \cdots \cdots \\ x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}=n\end{array}\right.$
Solution: Let $f(x)=(x-x_{1})(x-x_{2}) \cdots(x-x_{n})$, by Vieta's formulas we get $f(x)=x^{n}-\sigma_{1} x^{n-1}+\sigma_{2} x^{n-2}-\cdots+(-1)^{n} \sigma_{n}$, and $f\left(x_{i}\right)=0(1 \leqslant i \leqslant n)$. Therefore, $x_{i}^{n}-\sigma_{1} n_{i}^{n-1}+\sigma_{2} n_{i}^{n-2}-\cdots+(-1)^{n} \sigma_{n}=0(1 \leqslant i \leqslant n)$

Multiply the given $n$ equations by $(-1)^{n-1} \sigma_{n-1},(-1)^{n-2} \sigma_{n-2}, \cdots$, $-\sigma_{1}, 1$, respectively, then add $1+1+\cdots+1=n$ multiplied by $(-1)^{n} \sigma_{n}$. Using equation (1) and $1-\sigma_{1}+\sigma_{2}+\cdots+(-1)^{n} \sigma_{n}=f(1)$, we get $0=n f(1)$, from which it follows that 1 is a root of $f(x)$, let's assume $x_{n}=1$. Replacing $n$ with $n-1$, and applying the same method to $x_{1}, x_{2}, \cdots, x_{n-1}$, we have $x_{1}=x_{2}=\cdots=x_{n}=1$.