Proof: By Induction
(related to Lemma: A Criterion for Associates)
"$\Rightarrow$"
 Assume, $a,b\in R$ are associates $a\sim b.$
 By definition, $a\mid b$ and $b\mid a.$
 This means that there are $c,d\in R$ with $ac=b$ and $bd=a.$
 Therefore, $a=bd=acd$, which means $0=a(cd1)$
 Since $(R, + , \cdot)$ is an integral domain, we have either $a=0$ or $cd=1$
 If $a=0$ then $b=0$ and, trivially, $a=cb$ for some $c\in R^\ast,$ ($(R^\ast,\cdot)$ being the group of units).
 If $a\neq 0$ then $cd=1$ but then $c$ is a unit, therefore $c\in R^\ast,$ and again $a=cb.$
"$\Leftarrow$"
 Let $a=cb$ with $c$ being a unit.
 Therefore, $b\mid a.$
 But since $c$ has an multiplicative inverse $d\in R^\ast$ with $cd=1$, it follows that $b=1\cdot b=(cd)b=(cb)d=ad.$
 It follows that $a\mid b.$
 By dfinition, the elements $a,b$ are associates $a\sim b.$
∎
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References
Bibliography
 Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013