My attempt:
If we take $n$ to be finite, since $\mathbb Z/n\mathbb Z$ is cyclic, there is a distinct subgroup of order $d$ for each divisor $d$ of $n$.
Any subgroup of a cyclic group is cyclic. Any subgroup of order $d$ is isomorphic to $\mathbb Z/d\mathbb Z$, as it is a cyclic group of order $d$.
Cyclic groups are abelian, and $gh=hg$ leads to $gH = Hg$, so all subgroups are normal.
Thus we can view $\mathbb Z/d\mathbb Z$ as the kernel of some homomorphism from $\mathbb Z/n\mathbb Z$ to a group $H$. By the first isomorphism theorem, we have $(\mathbb Z/n\mathbb Z)/(\mathbb Z/d\mathbb Z)$ is isomorphic to the image of that homomorphism.
It is at this point I've become stuck, as I'm not sure what I can say about the image of this homomorphism. I feel like it should be $\mathbb Z/k\mathbb Z$, where $k$ is the integer such that $kd = n$, and this would prove the quotient group is cyclic, as it is isomorphic to a cyclic group. However, I am not sure if this is right or how to justify this?
Any direction is greatly appreciated!
A quotient group $H$ of a group $G$ has a surjective homomorphism $f:G\to H$. If $G$ is cyclic and $x$ is a generator of $G$, then every element of $G$ is $x^n$ for some $n$. Thus the elements of $H$ consist of $f(x^n) =f(x)^n$ since $f$ is surjective. Thus $f(x) $ generates $H$, so $H$ is cyclic.