For groups elements $a$ and $b$, how do you prove that $(a^{-1}ba)^n = a^{-1}b^na$? For any elements $a$ and $b$ from a group and any integer $n$, prove that $(a^{-1}ba)^n = a^{-1}b^na$.


Suppose $P(n):$ $(a^{-1}ba)^n = a^{-1}b^na$.

$1.$ $n=1$ $(a^{-1}ba)^1 =a^{-1}ba =a^{-1}b^1a$

$2.$ Suppose $P(n)$ is true we want to show $P(n+1)$ is true. 

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

Then using associative and inverse property this reduces to 
$a^{-1}b^{n+1}a$

Thus $P(n+1)$ is true, therefore $P(n)$ is true for every natural number. 

I'm not sure how to show it for the negative numbers or could I just use this result to conclude it holds for the negative numbers also.
I suggest that you prove
(for $n\geq 1$)
$(a^{-1} ba)^n (a^{-1} b^{-n} a)$
is equal to the neutral element of the group,
thus calculating the inverse of $(a^{-1} ba)^n$.

This way, you dont need to do another induction.
Induction is overkill. Just write it down:

$$
(a^{-1}ba)^n = 
\underbrace{(a^{-1}ba)(a^{-1}ba)\dotsb(a^{-1}ba)}_{n \text{ copies}} = 
\underbrace{a^{-1}baa^{-1}ba\dotsb a^{-1}ba}_{n \text{ copies}} = 
a^{-1}\underbrace{bb\dotsb b}_{n \text{ copies}}a = 
a^{-1}b^na
$$