15. 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 for $\left\{a_{n}\right\}$.
$$
\begin{array}{l}
\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1} \Rightarrow \sqrt{\frac{a_{n}}{a_{n-1}}}-1=2 \sqrt{\frac{a_{n-1}}{a_{n-2}}} \\
\Rightarrow \sqrt{\frac{a_{n}}{a_{n-1}}}+1=2\left(\sqrt{\frac{a_{n-1}}{a_{n-2}}}+1\right) \Rightarrow\left\{\sqrt{\frac{a_{n}}{a_{n-1}}}+1\right\} \text { is a geometric sequence with common ratio } 2 \\
\Rightarrow \sqrt{\frac{a_{n}}{a_{n-1}}}+1=2^{n} \Rightarrow \frac{a_{n}}{a_{n-1}}=\left(2^{n}-1\right)^{2} \Rightarrow a_{n}=\prod_{k=1}^{n}\left(2^{k}-1\right)^{2}(n \geqslant 2) . \\
\text { Therefore, } a_{n}=\left\{\begin{array}{l}
1, n=1, \\
\prod_{k=1}^{n}\left(2^{k}-1\right)^{2}, n \in \mathbf{N}, n \geqslant 2 .
\end{array}\right.
\end{array}
$$