= Solution
First establish the two general claims about <polycyclic groups>. Intersect a subnormal cyclic-factor series $(G_i)$ with $H\leq G$. Then $H\cap G_i\trianglelefteq H\cap G_{i+1}$, and the inclusion induces an <injective> homomorphism
$$
\frac{H\cap G_{i+1}}{H\cap G_i}\longrightarrow\frac{G_{i+1}}{G_i}.
$$
A subgroup of a <cyclic group> is cyclic, so this is a cyclic-factor <subnormal series> for $H$, after omitting repetitions. This proves <subgroups of polycyclic groups are polycyclic>.
To prove the generator bound, choose $g_i\in G_i$ whose coset generates $G_i/G_{i-1}$ whenever that factor is nontrivial. Inductively $G_i=\langle G_{i-1},g_i\rangle$, so $G=\langle g_1,\ldots,g_n\rangle$. This is the <generator bound from a cyclic-factor series>: \b[at most $n$ generators suffice].
For the first example, take $G=C_2^m$ with its coordinate flag
$$
0<C_2<C_2^2<\cdots<C_2^m.
$$
It has length exactly $m$ and each factor is $C_2$, so all factors are cyclic.
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