8. We prove by induction: When $1 \leqslant k \leqslant n$,
$$\sqrt{k a+\sqrt{(k+1) a+\cdots+\sqrt{n a}}}<1+\sqrt{k a}$$

When $k=n$, the above inequality clearly holds. Suppose $\sqrt{(k+1) a+\sqrt{(k+2) a+\cdots+\sqrt{n a}}}<1+$
$$\begin{array}{l} \sqrt{(k+1) a}, \text{ then } \sqrt{k a+\sqrt{(k+1) a+\cdots+\sqrt{n a}}}<\sqrt{k a+1+\sqrt{(k+1) a}}< \\ \sqrt{k a+1+2 \sqrt{k a}}=1+\sqrt{k a} . \end{array}$$

Therefore, (1) holds for all $k(1 \leqslant k \leqslant n)$. In particular, when $k=1$, the original inequality holds.