One. (25 points) Let the sequence $\left\{x_{n}\right\}$ satisfy
$$
x_{1}=1, x_{n}=\sqrt{x_{n-1}^{2}+x_{n-1}}+x_{n-1}(n \geqslant 2) \text {. }
$$

Find the general term formula for the sequence $\left\{x_{n}\right\}$.
(Zhang Lei, provided)
One can easily prove that $x_{n}$ is positive.
Notice that
$$
\begin{array}{l}
\frac{1}{x_{n}}=\frac{1}{\sqrt{x_{n-1}^{2}+x_{n-1}}+x_{n-1}} \\
=\frac{\sqrt{x_{n-1}^{2}+x_{n-1}}-x_{n-1}}{x_{n-1}}=\sqrt{1+\frac{1}{x_{n-1}}}-1,
\end{array}
$$

which means
$$
\begin{array}{l}
\frac{1}{x_{n}}+1=\sqrt{1+\frac{1}{x_{n-1}}}=\cdots \\
=\left(1+\frac{1}{x_{1}}\right)^{\frac{1}{2^{n-1}}}=2^{\frac{1}{2^{n-1}}}=2^{2^{1-n}} . \\
\text { Therefore, } x_{n}=\frac{1}{2^{2^{1-n}}-1} .
\end{array}
$$