Powers and Roots of Group Elements Let $G$ be a group, and $a$, $b \in G$ 


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*$(bab^{-1})^{n} = ba^{n}b^{-1}$, 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 :)

Both of the proofs look good to me.   
