Example 2 Calculate $[\sqrt{2008+\sqrt{2008+\cdots+\sqrt{2008}}}]$ (2008 appears 2008 times).
(2008, International Youth Math Invitational Competition)
【Analysis】Although there are 2008 square root operations, as long as you patiently estimate from the inside out, the pattern will naturally become apparent.
Let $a_{n}=\sqrt{2008+\sqrt{2008+\cdots+\sqrt{2008}}}$ (with $n$ 2008s, $n=1,2, \cdots$ ).
Given $44<\sqrt{2008}<45$, we have
$$
\begin{array}{l}
45^{2}<2008+44<2008+\sqrt{2008} \\
<2008+45<46^{2} .
\end{array}
$$

Thus, $45<\sqrt{2008+\sqrt{2008}}<46$, which means $45<a_{2}<46$.
Therefore, $45^{2}<2008+45<2008+a_{2}$
$$
<2008+46<46^{2} \text {. }
$$

Hence, $45<\sqrt{2008+a_{2}}<46$, which means $45<a_{3}<46$.
Similarly, $45<a_{4}<46, \cdots \cdots 45<a_{2008}<46$.
Therefore, $\left[a_{2008}\right]=45$.