Example 1 Let $\alpha=\frac{\sqrt{5}+1}{2}$. Then $\left[\alpha^{16}\right]=$ $\qquad$ .
(2008, "Five Sheep Cup" Mathematics Competition (Junior High School))
[Analysis] Note that $\alpha=\frac{\sqrt{5}+1}{2}$ and $\beta=\frac{\sqrt{5}-1}{2}$ $(0<\beta<1)$ can make $\alpha+\beta=\sqrt{5}$ and $\alpha \beta=1$. Therefore, the value of $\alpha^{16}+\beta^{16}$ can be calculated to make a judgment.
Let $\beta=\frac{\sqrt{5}-1}{2}$. Then
$$
\begin{array}{l}
\alpha+\beta=\sqrt{5}, \alpha \beta=1 . \\
\text { Therefore, } \alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=3, \\
\alpha^{4}+\beta^{4}=\left(\alpha^{2}+\beta^{2}\right)^{2}-2(\alpha \beta)^{2}=7, \\
\alpha^{8}+\beta^{8}=\left(\alpha^{4}+\beta^{4}\right)^{2}-2(\alpha \beta)^{4}=47, \\
\alpha^{16}+\beta^{16}=\left(\alpha^{8}+\beta^{8}\right)^{2}-2(\alpha \beta)^{8}=2207 .
\end{array}
$$

Since $0<\beta<1$, we know $0<\beta^{16}<1$.

Therefore, $\left[\alpha^{16}\right]=2206$.