> [!NOTE] Theorem (Equivalent definitions of negative power) > Let $G$ be a [[Groups|group]]. Let $a\in G.$ If $n\in\mathbb{Z}$ then the $(a^{-1})^{n}=(a^{n})^{-1} =a^{-n}$ where $a^{n}$ denotes the $n$th [[Integer Power of Group Element|power]] of $a.$ **Proof**: This clearly true if $n=0$ since all three expressions are equal to $1.$ If $n\in\mathbb{N}^{+},$ then $a^{n}(a^{-1})^{n}=\underbrace{aa\cdots a}_{n\mathrm{~times}} \; \underbrace{a^{-1}a^{-1}\cdots a^{-1}}_{n\mathrm{~times}}.$Each $aa^{-1}$ in the centre gives $1$ and eventually we see that this expression is $1.$ Therefore $(a^{-1})^{n}=(a^{n})^{-1}.$ Also $(a^{n})^{-1}=a^{-n}$ by definition. If $n\in \mathbb{Z}^{-},$ then $a^n(a^{-1})^n=(a^{-n})^{-1}((a^{-1})^{-n})^{-1}=(a^{-1})^{-n}((a^{-1})^{-1})^{-n}=(a^{-1})^{-n}a^{-n}=1$using the result for positive integers. Again the last equality is true by definition.