8、Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{n}=n(1 \leq n \leq 5), a_{n+1}=a_{1} \cdot a_{2} \cdot \ldots \cdot a_{n}-1(n \geq 5)$, then $S_{m}=a_{1} \cdot a_{2} \cdot \ldots \cdot a_{m}-a_{1}^{2}-a_{2}^{2}-\ldots-a_{m}^{2}$ takes the maximum value as $\qquad$

Given the sequence $\{a_n\}$ satisfies $a_n = n (1 \leq n \leq 5), a_{n+1} = a_1 \cdot a_2 \cdot \ldots \cdot a_n - 1 (n \geq 5)$, then $S_m = a_1 \cdot a_2 \cdot \ldots \cdot a_m - a_1^2 - a_2^2 - \ldots - a_m^2$ takes the maximum value as $\qquad$
8.65

Analysis: It is easy to know that $S_{1}=0, S_{2}=-3, S_{3}=-8, S_{4}=-6, S_{5}=65$, when $m \geq 5$,
$$
\begin{aligned}
S_{m+1}-S_{m}=a_{1} a_{2} \ldots a_{m+1} & -a_{1} a_{2} \ldots a_{m}-a_{m+1}^{2}=a_{1} a_{2} \ldots a_{m}\left(a_{m+1}-1\right)-a_{m+1}^{2} \\
& =\left(a_{m+1}+1\right)\left(a_{m+1}-1\right)-a_{m+1}^{2}=-1
\end{aligned}
$$

Therefore, when $m \geq 5$, $S_{m}=70-m$, thus the maximum term in the sequence $\left\{S_{m}\right\}$ is 65.