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 $$nb \text{ (mod m)} \text{ will be } \{0, 1, 2, \dots, (m-1)\}$$

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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?