I would prove this along the lines of J Rotman - An introduction to the theory of Groups, Graduate Texts in Mathematics, Springer-Verlag, 4th Edition. Pg:28
Observe that the converse of what you'd like to prove is quite a trivial issue.
For what you'd like to prove, J R has the following Recipe:
Lemma: Let $\phi$ be the Euler's totient function that counts the number of numbers less than $n$ and coprime to it, $ \sum_{d|n} \phi (d) = n$
Proof:
The proof of this fact follows from the following: Let G be a cyclic group of order $n$. I'll assume you can prove that there are $\phi(n)$ generators for this $G$ and in general $\phi(k)$ generators for a cyclic group of order $k$.
$G=\bigcup gen(C)$ where $gen(C)$ stands for generators of cyclic subgroups $C$ of G.(So $C$ ranges over all cyclic subgroups of $G$) And this union is a disjoint union, from the definition of a generator.
Note that the disjoint union holds for all $G$ and there is nothing sacrosanct about $G$ being cyclic, except from here: I claim now that, as the subgroups are unique for each divisor, the disjoint union translates into a sum that is exactly the lemma we seek to prove.
Your Question:
As pointed out before, I'll make use of the disjoint union.
$ n=|G| = \sum |gen(C)| \leq \sum_{d|n}{\phi(d)}$
So, now I use my lemma to conclude that, I have cyclic subgroups for each divisor of $n$ and in particular for $n$. So, that must be $G$. So, G must be cyclic.
I would like to emphasise, that this classifies all finite cyclic groups.