Proof

By hypothesis, $(R, + ,\cdot)$ is an integral domain.
Let $R^\ast\subset R$ be the subset of $R$ containing all units of $R.$
Obviously, $1\in R^\ast,$ therefore $R^\ast$ is not empty.
If $a\in R^\ast$ then, by definition of a unit, there exists an $b\in R$ with $a\cdot b=1.$
Since $R$ is an integral domain, the multiplication "$\cdot$" is commutative.
Therefore, $b\cdot a=1,$ thus $b$ is a also unit.
Therefore $b\in R^\ast,$ if $a\in R^\ast.$
We have shown that $R^\ast$ contains always both, an element $a\in R^\ast$ and also its multiplicative inverse $b\in R^\ast.$
Since "$\cdot$" is associative, we have shown all properties of a group.
This proves that the group of units $(R^\ast,\cdot)$ is indeed a group.
∎

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