Exercise 1.1.34 (If $|x| = \infty$, then the elements $x^n,\, n\in \mathbb{Z}$ are all distinct)

Proof. By definition, $x$ is an element of infinite order if and only if $x^k\ne 1$ for all integers $k>0$.

Suppose for contradiction that the elements $x^n$, $n\in \mathbb{Z}$ are not all distinct. Then there are some integers $n,m$ such that

Put $k=m-n>0$. Then $x^k=x^{m-n}=1$, but since $k>0$, $x^k\ne 1$ by definition. This contradiction shows that the elements $x^n$, $n\in \mathbb{Z}$, are all distinct. □