8. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{n+1}>a_{n}$, and $a_{n+1}^{2}+a_{n}^{2}+1=2\left(a_{n+1} a_{n}+a_{n+1}+a_{n}\right)$, then $a_{n}=$
8. $n^{2}$  From the given, we have $1+a_{n+1}^{2}+a_{n}^{2}+2 a_{n+1} a_{n}-2 a_{n+1}-2 a_{n}=4 a_{n+1} a_{n}$, which means $\quad\left(a_{n+1}+a_{n}-1\right)^{2}=$ $4 a_{n} a_{n+1}$.
Since $a_{1}=1, a_{n+1}>a_{n}$, it follows that $a_{n+1}+a_{n}-1>0$.
Thus, $a_{n+1}+a_{n}-1=2 \sqrt{a_{n} a_{n+1}}$ or $\quad\left(\sqrt{a_{n+1}}-\sqrt{a_{n}}\right)^{2}=1$, which means $\sqrt{a_{n+1}}-\sqrt{a_{n}}=1$.
Given $a_{1}=1$, we know that the arithmetic sequence $\left\{\sqrt{a_{n}}\right\}$ has the general term $\sqrt{a_{n}}=n$. Therefore, $a_{n}=n^{2}$.