Proof

By the existence of integer one, we have $x=x\cdot 1$ all $x\in\mathbb Z\quad ( * ).$
Suppose, $1^{\ast}$ is any (other) integer number, for which $x=x\cdot 1^{\ast}$ all $x\in\mathbb Z\quad ( * * ).$
By $( * )$, we have $1^{\ast}=1^{\ast}\cdot 1.$
Since the multiplication of integers is commutative, we get $1^{\ast}=1\cdot 1^{\ast}.$
By $( * * )$ we get $1^{\ast}=1.$
Thus, the integer number one $1$ is unique.
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