If $x^m=e$ has at most $m$ solutions for any $m\in \mathbb{N}$, then $G$ is cyclic

Fraleigh(7th ed) Sec10, Ex47. Let $G$ be a finite group. Show that if for each positive integer $m$ the number of solutions $x$ of the equation $x^m=e$ in $G$ is at most $m$, then $G$ is cyclic.

I tried it a few hours but I couldn't solve it. So I read the solution. But I narrowly understood the solution, and it's still unclear to me. How can I solve it?

The proof that I know of this fact goes as follows. Say that $|G|=n$ and for $1\leq d\leq n$ let $$ A_d=\{g\in G: g \text{ has order } d\}. $$ Then one proves:

(1) that $A_d=\emptyset$ if $d$ is not a divisor of $n$;

(2) that if $d\mid n$, then $|A_d|\leq\varphi(d)$.

This allows us to conclude because the chain of inequalities $n=|G|=\sum_{d\mid n}|A_d|\leq\sum_{d\mid n}\varphi(d)=n$ gives actual equality at every step and, in particular, $A_n\neq\emptyset$, i.e. $G$ admits a generator.

There are many ways to proceed with this problem, so I give one which I don't remember seeing in a text, though perhaps it uses a little more knowledge of group theory. We proceed by induction. The result is true if $|G| =p$ for some prime $p$, (and if $|G| =1$), so suppose that $|G| >1$ is not prime and the result is true for groups of order less than $|G|$. If $H$ is any proper subgroup of $G$, then $H$ is cyclic. If $H$ has order $m$, then $H$ contains $m$ solutions in $G$ of $x^m = 1$. But there are only m solutions of $x^m = 1$ in $G$, so $H = \{ x \in G: x^m = 1 \}$. Notice that if $x^m = 1$ then $(gxg^{-1})^m = 1$ for any $g \in G$. Hence $gHg^{-1} = H$ for any $g \in G$. Hence every subgroup of $G$ is normal. In particular, each Sylow $p$-subgroup of $G$ is normal in $G$, and $G$ is the direct product of its Sylow $p$-subgroups. If all Sylow $p$-subgroups of $G$ are proper, then all are cyclic by the induction hypothesis, and then $G$ itself is cyclic. Hence we may suppose that $G$ is a $p$-group for some prime $p$. Let $p^e$ be the largest order of an element of $G$. Then $G$ has a cyclic subgroup, say $N$, of order $p^e$, and $N \lhd G$ as above. But choose $x \in G$. Then $x$ has order $p^f$ for some non-negative integer $f$. If $f >e$, the choice of $e$ is contradicted. If $f \leq e$, then $x \in N$, since $x^{p^e} = 1$ and $N$ contains all solutions in $G$ of $y^{p^e} = 1$. Hence $N = G$ and $G$ is cyclic.