Example 8 Let $n$ be a given integer greater than 5, solve the system of equations
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=n+2, \\
x_{1}+2 x_{2}+\cdots+n x_{n}=2 n+2, \\
x_{1}+2^{2} x_{2}+\cdots+n^{2} x_{n}=n^{2}+n+4, \\
x_{1}+2^{3} x_{2}+\cdots+n^{3} x_{n}=n^{3}+n+8 .
\end{array}\right.
$$

where $x_{i} \geqslant 0, i=1,2, \cdots, n$.
Analyzing the problem, the number of equations is less than the number of unknowns, and $x_{i} \geqslant 0$, so we can consider using the conditions for equality in inequalities to solve it.
Solving (2) -(1), (3)-(2), (4) -(3) we get:
$$
\left\{\begin{array}{l}
x_{2}+2 x_{3}+\cdots+(n-1) x_{n}=n, \\
2 x_{2}+2 \cdot 3 x_{3}+\cdots+n(n-1) x_{n}=n^{2}-n+2, \\
4 x_{2}+3^{2} \cdot 2 x_{3}+\cdots+n^{2}(n-1) x_{n}=n^{3}-n^{2}+4 .
\end{array}\right.
$$
(6) $-2 \times(5),(7)-2 \times(6)$ we get:
(8) n-(9) we get: $\sum_{i=3}^{\frac{n-1}{2}}(n-i)(i-1)(i-2) x_{i}=0$ Since $x_{1} \geqslant 0$, it follows that $x_{i}=0(i=3,4, \cdots, n-1)$ Substituting into (1),(2),(3) $\Rightarrow x_{1}=n, x_{2}=x_{n}=1$.
/Comment Using inequalities to solve equations of special forms is a common method.