Question: Let $G$ be a group and let $a \in G$.
Prove that $\left \langle a^{-1} \right \rangle=\left \langle a \right \rangle$
Suppose $\left \langle a^{-1} \right \rangle$ so $\left \langle a^{-1} \right \rangle=\left \{ a^{0},a^{-1},a^{-2},a^{-3},\cdot \cdot \cdot \right \}$
But $a \in G$ so by the inverse property $a^{-1} \in G \,\, \forall \, \, a \in G$
Thus,
$\left \langle a^{-1} \right \rangle=\left \{ \left ( a^{-1} \right )^{0},\left ( a^{-1} \right )^{-1},\left ( a^{-1} \right )^{-2},\left ( a^{-1} \right )^{-3},\cdot \cdot \cdot \right \} =\left \{ a^{0},a^{1},a^{2},a^{3},\cdot \cdot \cdot \right \} =\left \langle a \right \rangle$
as required
Note:$a^{0}=e$
Is my proof valid? Thanks in advance.