Example 7 If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$ .
Solution: Since $m \neq n$, by the definition of roots, $m, n$ are two distinct roots of the equation $x^{2}-x-1=0$. By Vieta's formulas, we have
$$
\begin{array}{l}
m+n=1, m n=-1 . \\
\because m^{2}+n^{2}=(m+n)^{2}-2 m n \\
\quad=1^{2}-2 \times(-1)=3, \\
\quad m^{3}+n^{3}=(m+n)^{3}-3 m n(m+n) \\
\quad=1^{3}-3 \times(-1) \times 1=4, \\
\therefore m^{5}+n^{5} \\
=-\left(m^{3}+n^{3}\right)\left(m^{2}+n^{2}\right)-(m n)^{2}(m+n) \\
\quad=4 \times 3-(-1)^{2} \times 1=11 .
\end{array}
$$