Five, let the sequence of positive numbers $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfy
$$
\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1} \quad(n \geqslant 2) .
$$

and $a_{0}=a_{1}=1$. Find the general term formula for $\left\{a_{n}\right\}$.
Five, transform $\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}$ into
$$
\sqrt{a_{n} a_{n-2}}-2 a_{n-1}=\sqrt{a_{n-1} a_{n-2}},
$$

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

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