Induction with compositions Proposition. Suppose    $g,h:\mathbb{R}\rightarrow\mathbb{R}, (g\circ h\circ g^{-1})^{n}=g\circ h^{n}\circ g^{-1}$ where  $n\in\mathbb{N}$ and $g$ is a bijection. 
We will prove this by mathematical induction.
We will prove the basis step $P(1)$ , that is $P(n)=
 (g\circ h\circ g^{-1})^{n}=g\circ h^{n}\circ g^{-1}$
We see that, $ P(1)=(g\circ h\circ g^{-1})^{1}.$
It is clear that $P(1)$
  is true because the composition is not being altered in any way.
Assume that $P(k)$
  is true. That is, $P(k)=(g\circ h\circ g^{-1})^{k}=g\circ h^{k}\circ g^{-1}.$
We need to show $P(k+1)$ is true. We see that: 
$P(k+1)=P(k)\circ(g\circ h\circ g^{-1})=(g\circ h^{k}\circ g^{-1})\circ(g\circ h\circ g^{-1}).$
I am have trouble understanding how $(g\circ h^{k}\circ g^{-1})\circ(g\circ h\circ g^{-1})=g\circ h^{k+1}\circ g^{-1}$.
Also, how does the rest of the proof look?

Looks like you're on the right track. Remember that function composition is associative and $g \circ g^{-1}$ is simply the identity function since g is bijective.Playing with the parentheses we have $(g \circ h^k) \circ (g^{-1} \circ g) \circ (h \circ g^{-1})=(g \circ h^k) \circ (h \circ g^{-1})$ which is equal too...
