Inverse of Group Commutator

Theorem
Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$.

Then $\sqbrk {g, h}$ is the inverse of $\sqbrk {h, g}$:
 * $\sqbrk {g, h} = \sqbrk {h, g}^{-1}$