Equivalence of Definitions of Order of Group Element

Theorem
Let $G$ be a group whose identity is $e$.

Let $a \in G$ have finite order such that $\left|{a}\right| = k$.

Then:
 * $\left|{\left \langle {a} \right \rangle}\right| = k$

where $\left \langle {a} \right \rangle$ is the smallest subgroup of $G$ containing $a$.

That is, the order of the subgroup generated by $a$ is equal to the order of $a$.

(Some sources use this as the definition of the order of an element, and from it derive the definition of the order of a group element.)

Proof
It follows straight away from List of Elements in Finite Cyclic Group that $\left|{\left \langle {a} \right \rangle}\right| = k$:
 * $\left \langle {a} \right \rangle = \left\{{a^0, a^1, a^2, \ldots, a^{k - 1}}\right\}$