Proving $(x^2)^{-1} = (x^{-1})^2$ in context of group theory. How would I go about proving  $(x^2)^{-1} = (x^{-1})^2$ in terms of group theory?
Would I start by multiplying both sides by the inverses of the LHS and RHS and using the property of the identity? Or should I use induction?
Thanks.
 A: By $(x^2)^{-1}$ you mean the inverse of $x^2$.  So, you can just show that $(x^2)\cdot (x^{-1})^2$ is the identity.
A: 
Note: $$(x^2)^{-1} = (x\cdot x)^{-1} = (x^{-1}\cdot x^{-1}) = (x^{-1})^2\tag{*}$$

So we have

$$(x^2)^{-1} = (x^{-1})^2.$$

as desired.

Added, for clarification, re comments:
$$(x^2)^{-1} = (x^{-1})^2$$ $$\iff$$ $$(x\cdot x)^{-1} = (x^{-1}\cdot x^{-1})\tag{1}$$
$$\iff$$ $$x^{-1}\cdot x^{-1} = x^{-1}\cdot x^{-1}\tag{2}$$
$(2)$ can certainly be reduced to look "prettier" by, say, left-multiplying by $x$, or by group cancellation laws, to arrive at the obvious equivalence: $e = e$, where $e$ is the identity. 
But equation $(*)$ at the top is more straightforward and direct for establishing equality. Note that $(*)$ uses the fact that $(a\cdot b)^{-1} = (b^{-1}\cdot a^{-1})$, which you've already proven, according to your comment below. In this case $a = b = x$.
A: Hint: $(a\cdot b)^{-1}=b^{-1}\cdot a^{-1}$, and $x^2=x\cdot x$.
A: Use the rule that $(ab)^{-1} = b^{-1} a^{-1}$.
To prove this rule, start with $(ab)^{-1} ab = e$ and then successively right-multiply by $b^{-1}$ and then $a^{-1}$.
