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Proof: By Induction

(related to Proposition: Principal Ideal Generated by A Unit)

"$\Rightarrow$"

Let $c$ is a unit of an integral domain $(R, + ,\cdot).$
This means, there is a multiplicative inverse $b\in R$ with $bc=1.$
Since $bc\in ( c ),$ we have $1\in ( c ).$
Therefore $( c )=R.$

"$\Leftarrow$"

Let $( c )=R.$
In particular, $1\in ( c ).$
Therefore, there is an $b\in R$ with $bc=1.$
Therefore, $c\in R^\ast.$
∎

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References

Bibliography

Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013