Show every submodule of $Z_{p^{\infty}}$ is of the form $(\frac{1}{p^{n}})\mathbb{Z}/\mathbb{Z}$ 
Let 
  $$Z_{p^{\infty}} = \{ x \in \mathbb{Q}/\mathbb{Z}: \exists n \in \mathbb{N} \text{ s.t. }  p^{n}x=0 \}$$
  be a $\mathbb{Z}$-submodule of $\mathbb{Q}/\mathbb{Z}$. We want to show every $\mathbb{Z}$-submodule of $M = Z_{p^{\infty}}$ is of the form $(\frac{1}{p^{n}})\mathbb{Z}/\mathbb{Z}$ for some $n \in \mathbb{N}$.

So we want to find for each submodule $N$ of $M$ an $n \in \mathbb{N}$ such that $N= (\frac{1}{p^{n}})\mathbb{Z}/\mathbb{Z}$.
I am struggling to see how we can find such an $n$. I can see that an element of $Z_{p^{\infty}}$ is an equivalence class of rational numbers of the form $\frac{a}{p^{n_{a}}}$ for some $n_{a} \in \mathbb{N}$ dependent on $a\in \mathbb{Z}$. But I'm struggling to find an argument as to why we can find an $n \in \mathbb{N}$ such that $p^{n}x = 0$ for all $x \in N$. Could someone help please?
EDIT: Essentially I can't see what's stopping elements of $N$ being of the form $\frac{a}{p^{n}}$ and $\frac{b}{p^{m}}$ where $\gcd(a,p)=\gcd(b,p)=1$ but $m\neq n$.

Consider the set $S=\{n\in\mathbb N:p^{-n}+\mathbb Z\in N\}$.
If $S$ is unbounded, then $p^{-n}+\mathbb Z\in N$ for each $n\in\mathbb N$, hence $N=Z_{p^\infty}$.
If $S$ is bounded let $n=\max S$, so that $p^{-n}+\mathbb Z\in N$ and $p^{-n-1}+\mathbb Z\not\in N$.
Then $N=\mathbb Zp^{-n}/\mathbb Z$.
