Im trying to prove that $\Bbb Z_n:=\Bbb Z/n\Bbb Z$ have cardinality $n$ just using properties of rings, this mean that Im trying to do it without the use of any multiplicative inverse $z^{-1}\in\Bbb Q$.

My try: if $|\Bbb Z_n|=n$ then must exist a bijective function of the kind

$$f:\{1,\ldots,n\}\to \Bbb Z_n$$

The easiest candidate is $f$ such that $f(k)=[k]$, where $[k]$ is the equivalence class of $k$, defined by

$$[k]=k+n\Bbb Z$$

Then two elements of $a,b\in\{1,\ldots,n\}$ are equivalent if

$$a\sim b\iff a\in b+n\Bbb Z\iff a-b\in n\Bbb Z$$

Then I can show that $f$ is injective, i.e. if $a,b\in\{1,\ldots,n\}$ and $a\neq b$ then $[a]\neq[b]$ due to the fact that

$$(a-b\in n\Bbb Z\iff b-a\in n\Bbb Z)\iff (|a-b|> 0\implies |a-b|\in n\Bbb N_{>0})$$

Then if $|a-b|\in n\Bbb N_{>0}$ we have that

$$|a-b|\le nk,\quad(\forall k\in\Bbb N_{>0})\land(\forall a,b\in\{1,\ldots,n\})$$

Because $\Bbb N_{>0}$ is well ordered to prove the last statement is enough to show that

$$|a-b|<n=\min(n\Bbb N_{>0})$$

(I dont will write this proof here, it is unnecessary to me at this point).

**The questions:** the problem that I have is that I dont know how to show that $f$ is surjective if I restrict myself to not use things like $z^{-1}\in\Bbb Q$ or any division rule. So,

there is an easy way to show the surjectivity of $f$? In fact,

there is an easier way to prove that $|\Bbb Z_n|=n$?

Thank you in advance!