8 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=0, a_{n+1}=a_{n}+1+2 \sqrt{1+a_{n}}(n=1$, $2, \cdots)$, then $a_{n}=$ $\qquad$
8 From the given, we have
$$
a_{n+1}+1=a_{n}+1+2 \sqrt{1+a_{n}}+1=\left(\sqrt{a_{n}+1}+1\right)^{2},
$$

and $a_{n}+1>0$. Therefore,
$$
\sqrt{a_{n+1}+1}=\sqrt{a_{n}+1}+1,
$$

which means $\left\{\sqrt{a_{n}+1}\right\}$ is an arithmetic sequence with the first term and common difference both equal to 1, so $\sqrt{a_{n}+1}=n$, i.e., $a_{n}=n^{2}-1$.