Prove if $(a,p)=1$, then $\{a,2a,3a,...,pa\}$ is a complete residue system modulo $p$. let $p$ be a prime and let $a$ be an integer not divisible by $p$, that is $(a,p)=1$. Then $\{a,2a,3a,...,pa\}$ is a complete residue system modulo $p$.
Since $p\nmid a$, $p\nmid\{a,2a,3a,...,(p-1)a\}$
Since $p|p$, $p|\{pa\}$ which is exactly one element in the set and is therefore a complete residue system. 
Wondering if this is an acceptable proof to this theorem. 
 A: $$k_1a \equiv k_2a \pmod p \iff p \ | \ (k_1 - k_2)a \iff p \ | \ (k_1 - k_2)$$
Since in our case $k_1, k_2 \in A = \{1, 2, ..., p\}$ we have that $\left| { k_1 - k_2} \right| \lt p $ and $p$ will not divide $(k_1 - k_2)$ if both elements are from $A$. So for any two elements $r, q \in A \;\; rp \not \equiv qp \pmod p $. Which says that every element in $\{a,2a,3a,...,pa\}$ is not congruent to each other modulo $p$. Then each of them must be distinct modulo $p$. Since we have $p$ elements in the set it forms a complete set of residues modulo $p$. 
A: ${\rm mod}\ p\!:\ f(x) = x^p\!-x \equiv(x\!-\!1)(x\!-\!2)\cdots(x\!-\!p)\, $ by $\,f\,$ has roots $\,1,\ldots,p,\,$ by little Fermat.   
Thus $\ \ (a^{-1}x)^p\!-a^{-1}x \equiv (a^{-1}x\!-\!1)(a^{-1}x\!-\!2)\cdots(a^{-1}x\!-\!p)\,\ $ by $\,\ x\mapsto a^{-1}x$ 
Thus we infer $\ \ x^p-x\equiv (x\!-\!a)(x\!-\!2a)\cdots(x\!-\!pa)\ $ by multiply LHS by $\,a,\,$  RHS by $\ a^p\equiv a$
Therefore $ \{1,2,\ldots,p\}\equiv \{a,2a,\ldots,pa\},\,$ both being all roots of $\,f(x)\,$ in the field $\,\Bbb Z/p$.
