Example 11 Given $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$. Then $m^{5}+n^{5}=$ $\qquad$ .
(From the riverbed Jiangsu Province Junior High School Mathematics Competition)
Solution: Construct a quadratic equation $x^{2}-x-1=0$. From the given conditions, we know that $m$ and $n$ are the two roots of the equation $x^{2}-x-1=0$, so $m+n=1, m n=-1$. Then
$$
\begin{array}{l}
m^{2}+n^{2}=(m+n)^{2}-2 m n=3 . \\
m^{4}+n^{4}=\left(m^{2}+n^{2}\right)^{2}-2 m^{2} n^{2}=7 \text {. } \\
\therefore m^{5}+n^{5} \\
=(m+n)\left(m^{4}+n^{4}\right)-m n(m+n) \\
\text { - }\left[(m+n)^{2}-3 m n\right] \\
=7-(-1) \times\left[1^{2}-3 \times(-1)\right]=11 \text {. } \\
\end{array}
$$