The lemma to be proved is:
Let $G$ be a finite abelian $p$-group that is not cyclic. Suppose that $g \in G$ has maximal order. If $h \in G \setminus \langle g \rangle$ has smallest possible order, then $|h| = p$.
And the proof is written as:
Let $g \in G$ be of maximal order in $G$, say $|g| = p^m$ for some $m \le n$. Since $G$ is not cyclic, $G \neq \langle g \rangle$. Choose $h \in G \setminus \langle g \rangle$ where $h$ has smallest possible order, say $|h|=p^l$. Since $e \in \langle g \rangle$, then $h \neq e$ and so $l > 0$. But $|h^p|=p^{l-1}$ and so $|h^p|$ has smaller order than $|h|$, whence $h^p \in \langle g \rangle$.
What I could not understand is the bolded line: how can we say that $h^p \in \langle g \rangle$ because $|h^p|$ has smaller order than $|h|$?
We chose $h$ to be an element of $G \setminus \langle g\rangle$ whose order is as small as possible.
Therefore if $h^p$ has smaller order than $h$, $h^p$ cannot be an element of $G \setminus \langle g\rangle$: otherwise, we would have chosen $h^p$ instead of $h$. This means it must be an element of $\langle g\rangle$.