Proof
(related to Lemma: Cyclic Groups are Abelian)
 Let \((G,\ast)\) be a cyclic group generated by the element \(g\), i.e. \(G=\langle g \rangle\).
 Then any two elements \(x,y\in G\) can be represented as \(x=g^k\) and \(y=g^l\) for some integers \(k,l\in\mathbb Z\).
 It follows by the exponentiation in a group and commutativity of integer addition that \[\begin{array}{rcl}x\ast y&=&g^k\ast g^l\\&=&g^{k+l}\\&=&g^{l+k}\\&=&g^l\ast g^k\\&=&y\ast x,\end{array}\]
which means that \(G\) is also Abelian.
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References
Bibliography
 Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013