Proving that $nb \text{ (mod m)}$ reaches all values $\{0 \dots (m-1)\}$ if $n$ and $m$ are relatively prime I am trying to prove the frobenius coin problem which requires me to prove the following lemma:
If $n$ and $m$ are relatively prime and $b$ is any integer, then the set of all possible values of $$nb \text{ (mod m)} \text{ will be } \{0, 1, 2, \dots, (m-1)\}$$

I have so far that the set of all values of
$b \text{ (mod m)}$ for any $b$ will be $\{0, \dots, (m-1)\}$, and that 
$$nb \text{ (mod m)} = [n \text{ (mod m)}\cdot b \text{ (mod m)]}\text{ (mod m)}$$
Let $n \text{ (mod m)} = p$. Then I need to prove that when the elements of the set $\{0, p, 2p, 3p, \dots, (m-1)p\}$ are reduced modulo $m$, then the result is just $$\{0, 1, 2, \dots, (m-1)\}$$ However, I am unsure how to do this. Can someone give me a hint?

By Euclid, $\,(m,n)=1,\ m\mid nx\,\Rightarrow\, m\mid x,\,$ i.e. $\,nx\equiv 0\,\Rightarrow x\equiv 0\,$ in $\,\Bbb Z/m,$
i.e. the linear map $\,x\mapsto nx\,$ has trivial kernel, thus it is $\,\color{#c00}1$-$\color{#c00}1\,$ on $\,\Bbb Z/m.$
But $\,\Bbb Z/m\,$ is finite, thus, by $\rm\color{#c00}{pigeonhole}$, the map is $\color{#c00}{\rm onto}$, proving the claim.
