Prove $(bab^{-1})^n = ba^nb^{-1}$ by induction

Question

p(1): $(bab^{-1})^1 = ba^1b^{-1}.$

p(k + 1): $(bab^{-1})^{k + 1} = (bab^{-1})^k(bab^{-1}) = b^ka^kb^{-k}bab^{-1} = ba^{k + 1}b^{-1}.$

Would that work?

Answer 1

You have the majority of the argument, with a presumed typo, so the rest becomes trying to write it with proper grammar and structure.

Claim: $(bab^{-1})^n = ba^n b^{-1}$ for all $n\in\mathbb{N}$

Base case: let $n=1$. Then $(bab^{-1})^1 = bab^{-1}$

Assume for our induction hypothesis that $(bab^{-1})^n = ba^n b^{-1}$ for some $n\geq 1$.

We show then that it must also be true for $n+1$.

$(bab^{-1})^{n+1} = (bab^{-1})^n(bab^{-1}) =^{I.H.} ba^nb^{-1}bab^{-1}$

$=ba^n(b^{-1}b)ab^{-1} = ba^n 1 a b^{-1} = ba^nab^{-1}=ba^{n+1}b^{-1}$

Thus, proving the claim.

Answer 2

Hint $\ $ The inductive step arises by scaling both sides by $\,bab^{-1}$ then removing $\,bb^{-1}$ on RHS.