Let $G$ be a group, and $a$, $b \in G$
$(bab^{-1})^{n} = ba^{n}b^{-1}$, for every positive integer $n$ \begin{align*} \text{Let P(n) be the statement: } (bab^{-1})^{n} &= ba^{n}b^{-1} \newline \text{Show the base case P(1) : } bab^{-1} &= bab^{-1} \newline \text{Assume P(k) is true : } (bab^{-1})^{k} &= ba^{k}b^{-1} \newline \text{Now I need to show P(k+1) is true by multiplying P(k) by } bab^{-1} \newline (bab^{-1})^{k}(bab^{-1}) &= ba^{k}b^{-1}(bab^{-1}) \newline &= ba^{k}(b^{-1}b)ab^{-1} \newline &= ba^{k}ab^{-1} \newline &= ba^{k+1}b^{-1} \newline \text{Which is the statement P(k+1)} \end{align*}
If $a^{-1}$ has a cube root, so does $a$.
If $a^{-1}$ does have a cube root, then there is an element $x$ in $G$ such that $a^{-1} = x^{3}$. \begin{align*} a^{-1} &= x^{3} \newline a^{-1}a &= x^{3}a \newline e &= x^{3}a \newline (x^{-1})^{3} &= (x^{-1})^{3}x^{3}a \newline (x^{-1})^{3} &= a \end{align*}
Comments on the correctness or ways to improve either proof would be appreciated :)