13. Let the sequence of positive numbers $a_{0}, a_{1}, a_{2}, \cdots$, 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 this sequence.
13. Using the recursive formula, we get $a_{0}=1, a_{1}=1=\left(2^{1}-1\right)^{2}, a_{2}=9=\left(2^{2}-1\right)^{2} \cdot\left(2^{1}-\right.$ $1)^{2}$, 
$a_{3}=21^{2}=\left(2^{3}-1\right)^{2}\left(2^{2}-1\right)^{2}\left(2^{1}-1\right)^{2}$, thus we conjecture: when $n \in \mathrm{N}$, we have $a_{n}=\left(2^{1}-1\right)^{2} \cdot\left(2^{2}-1\right)^{2} \cdots \cdot\left(2^{n}-1\right)^{2}$.
From $\sqrt{a_{k+1} a_{k-1}}-\sqrt{a_{k} a_{k-1}}=2 a_{k}$ we get
$$
a_{k+1}=\left[\sqrt{a_{k}}+\frac{2 a_{k}}{\sqrt{a_{k-1}}}\right]^{2}=\left(2^{1}-1\right)^{2}\left(2^{2}-1\right)^{2} \cdots\left(2^{k+1}-1\right)^{2} \text {, }
$$

This leads to $a_{n}=\left(2^{1}-1\right)^{2} \cdots\left(2^{n}-1\right)^{2}$.