Proving that in a Group the inverse of the inverse of an element is the element itself
Solution 1:
$(a^{-1})^{-1} = a$ because $a^{-1} * a = a *a^{-1} = e$. Hence $a^{-1}*(a^{-1})^{-1} = a^{-1} * a = e$ etc.
Maybe this makes it more clear: Write $b:=a^{-1}$. Then we try to prove that $b*a = a*b = e$ so that $b^{-1} = a$.