8. (1) 数列 $\left\{a_{n}\right\}$ 满足: $a_{1}=1$, 且 $a_{n+1}-a_{n}=2^{n}+1$, 求 $a_{n}$.

(2) 已知数列 $\left\{a_{n}\right\}$ 满足: $a_{1}=1$, 且 $n a_{n+1}=(n+1) a_{n}$, 求 $a_{n}$
(1) $a_{n+1}-a_{n}=2^{n}+1$

$a_{n}-a_{n-1}=2^{n-1}+1$

$a_{2}-a_{1}=2^{1}+1$

累加可得: $a_{n}-a_{1}=2+2^{2}+\cdots+2^{n-1}+(n-1)$

$$
=\frac{2\left(2^{n-1}-1\right)}{2-1}+n-1=2^{n}+n-3
$$

$\therefore a_{n}=2^{n}+n-2$

(2) $n a_{n+1}=(n+1) a_{n} \Rightarrow \frac{a_{n+1}}{a_{n}}=\frac{n+1}{n}$

$\therefore \frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_{2}}{a_{1}}=\frac{n}{n-1} \cdot \frac{n-1}{n-2} \cdots \cdots \frac{2}{1}$

$\Rightarrow \frac{a_{n}}{a_{1}}=n \quad \therefore a_{n}=n a_{1}=n$