Five, let the sequence of positive numbers $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfy $\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}(n \geqslant 2)$, and $a_{0}=a_{1}=1$, find the general term formula of $\left\{a_{n}\right\}$.
$$
a_{n}=\left[\begin{array}{l}
1 \quad(n=0), \\
\prod_{k=1}^{n}(2 k-1)^{2}(n \in \mathbf{N})
\end{array}\right.
$$

Five, 【Analysis and Solution】Transform $\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}$ to
$$
\sqrt{a_{n} a_{n-2}}-2 a_{n-1}=\sqrt{a_{n-1} a_{n-2}},
$$

Divide both sides by $\sqrt{a_{n-1} a_{n-2}}$ to get
$$
\sqrt{\frac{a_{n}}{a_{n-1}}}-2 \sqrt{\frac{a_{n-1}}{a_{n-2}}}=1
$$

Thus, $\sqrt{\frac{a_{n-1}}{a_{n-2}}}-2 \sqrt{\frac{a_{n-2}}{a_{n-3}}}=1$
$$
\sqrt{\frac{a_{2}}{a_{1}}}-2 \sqrt{\frac{a_{1}}{a_{0}}}=1
$$
(1) $\times 1+$ (2) $\times 2+$ (3) $\times 2^{2}+\cdots+$ (n-1) $\times 2^{n-2}$,

We get $\sqrt{\frac{a_{n}}{a_{n-1}}}-2^{n-1} \sqrt{\frac{a_{1}}{a_{0}}}=1+2+2^{2}+\cdots+2^{n-2}$,
Thus, $\sqrt{\frac{a_{n}}{a_{n-1}}}=1+2+2^{2}+\cdots+2^{n-1}=2^{n}-1$.
$$
\begin{array}{l}
\therefore a_{n}=\left(2^{n}-1\right)^{2} a_{n-1}\left(2^{n}-1\right)^{2}\left(2^{n-1}-1\right)^{2} a_{n-2}=\cdots=\prod_{k-1}^{n}\left(2^{k}-1\right)^{2}, \\
\therefore a_{n}=\left[\begin{array}{ll}
1 & (n=0) \\
\prod_{k=1}^{n}(2 k-1)^{2} & (n \in \mathbf{N})^{.}
\end{array}\right.
\end{array}
$$