I understand everything except for the part where the author verifies whether
\begin{equation}\langle a\rangle = \{e, a^1, a^2, \ldots, a^{n-1}\}.\end{equation}
To me this seems completely superfluous. It is the definition of a group generated by $a$. Based on the line "Certainly..." I suppose his argument is that we cannot be certain that $\langle a\rangle$ contains only the elements above, perhaps it contains more. Then he proves it does not contain more elements.
But why? Why can we not be certain that it does not contain more elements?
No, it's not superfluous: the definition of the subgroup generated by $a$ gives $\langle a\rangle=\{a^n: n\in\mathbb Z\}$. When $a$ has order $1\le n<\infty$ you must prove that $\langle a\rangle = \{e, a^1, a^2, \ldots, a^{n-1}\}$. Obviously (or "certainly") the definition of $\langle a\rangle$ shows that $\{e, a^1, a^2, \ldots, a^{n-1}\}\subseteq\langle a\rangle $, and now you have to prove the converse. Take an $m\in\mathbb Z$ and prove that $a^m\in\{e, a^1, a^2, \ldots, a^{n-1}\}$: write $m=nq+r$ with $0\le r<n$ and then $a^m=(a^n)^qa^r=a^r$ which clearly belongs to $\{e, a^1, a^2, \ldots, a^{n-1}\}$.