Show that if $m$ is prime then the product of two non-zero elements of $\mathbb Z/m\mathbb Z$ is again non-zero I am having some trouble concluding this. 

I have the following:

$[a][b]=[ab]$ by the definition of multiplication in $\mathbb Z/m\mathbb Z$

By the remainder theorem, we know that $a=qm+r$ and $b=km+s$ where $q,k,r,s\  \epsilon\  \mathbb Z$

Additionally,  $0\lt r\lt m$ and  $0\lt s\lt m$ 

*It is greater than (not greater than or equal to) because at the onset of this problem, we have defined $[a]$ and $[b]$ to be non-zero

So, with those stated:

$[a][b]=[qm+r][km+s]=[(qm+r)(km+s)]$ 

which is equal to $[m^2qk+qms+mkr+rs]$ 

which is simply equal to $[rs]$

This is where I run into trouble. I am aware that I need to show that $[rs]$ is not equal to zero (i.e. $r*s$ should not be a divisor of $m$ or some multiple of $m$...say $\alpha m$.)

I thought that I could maybe setup a contradiction by stating something like:

"Assume $r*s=\alpha* m$" and then go on to show this can only be the case if $r$ or $s$ is not coprime with m (we know that $r$ and $s$ must be coprime with $m$ if $m$ is prime because $r$ and $s$ are both less than $m$ and greater than $0$...which would be a contradiction!)

Unfortunately, I cannot figure out how to explicitly demonstrate this. Any help would be appreciated! 

edit: Using the recommended Euclid's Lemma, while retaining the structure of my argument, we have the following:

Assume $r*s=\alpha* m$

This means that $m|r*s$. Recall, $m$ is prime and therefore, by Euclid's Lemma, we have either $m|r$ or $m|s$. This produces a contradiction for the following reason:

$r$ and $s$ have both been defined to be between $0$ and $m$. A number cannot evenly divide a number that is smaller than itself but greater than $0$.

Therefore, $rs\neq \alpha*m$ which means that $[rs] \neq[0]$ (i.e. it is non-zero).
If $[rs]$ is zero then $m \mid rs$ but $m$ is prime so $m \mid r$ or $m \mid s$ (no loss of generality) let's assume that $m\mid r$ then $m \mid a = qm +r$, i. e., $[a]$ is zero, wich is a contradition.
The answer is much easier. 

Take $~x~$ a non multiple of $~m~$. Then the additive cyclic group generated by $~[x]~$ is cyclic of a order dividing $~m~,~$ thus $~m~$ since $~m~$ is prime. 

Therefore $~n[x]=[nx]=0=[m]~$ if and only if $~n~$ multiple of $~m~$.