Proof

By hypothesis, $(G,\ast)$ is a group and $a\in G$ its element with a finite order $\operatorname{ord}(a)=n < \infty.$
By definition of generating set of a group, we have $\langle a\rangle=\{a^k\mid k\in\mathbb Z\}.$
By division with quotient and remainder, we can write $k=qn+r$ with $0\ge r < n.$
Because $n$ is the order of the element $a$, we have $a^{k}=a^{qn+r}=(a^n)^q\cdot a^r=e^q\cdot a^r=a^r$ with $e\in G$ being the neutral element.
Therefore, $\langle a\rangle=\{a^0,a^1,a^2,\ldots a^{n-1}\}.$
Moreover, all of these elements are distinct.
- Assume, they were not, e.g. $a^i=a^j$ but $0\le i < j\le n-1.$
- Then, we would have $a^{j-i}=e,$ in contradiction to $0 < j-i < n.$
- Therefore, $n=\operatorname{ord}(a)$ equals the group order $|\langle a\rangle|.$
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