Proof
(related to Proposition: Finite Order of an Element Equals Order Of Generated Group)
 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, $\langle a\rangle=\{a^0,a^1,a^2,\ldots a^{n1}\}.$
 Moreover, all of these elements are distinct.
 Assume, they were not, e.g. $a^i=a^j$ but $0\le i < j\le n1.$
 Then, we would have $a^{ji}=e,$ in contradiction to $0 < ji < n.$
 Therefore, $n=\operatorname{ord}(a)$ equals the group order $\langle a\rangle.$
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

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