Solution
= Solution
Apply (iii) successively to the <powerful p-groups> $G_i$. Starting with $G_1=\langle a_1,\ldots,a_n\rangle$, if $G_i=\langle a_1^{p^{i-1}},\ldots,a_n^{p^{i-1}}\rangle$, then
$$
G_{i+1}=G_i^p=\langle(a_1^{p^{i-1}})^p,\ldots,(a_n^{p^{i-1}})^p\rangle.
$$
Thus
$$
\boxed{G_{i+1}=\langle a_1^{p^i},\ldots,a_n^{p^i}\rangle\quad(i\geq1).}
$$