Suppose that $f:(a,b)\rightarrow$ is differentiable and that $f'$ is never zero. By the inverse-function theorrem, $f$ is a bijection from $(a,b)$ onto an interval $(c,d)$, the function $f^{-1}:(c,d)\rightarrow(a,b)$ is differentiable, and
$(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}, y\in(c,d)$.
Use this formula to show that, for all $r\in\mathbb{N}$, if $f$ is of class $C^r$ then $f^{-1}$ is of class $C^r$.
Things I know:
A function is of class $C^r$ if $f^{(r)}$ exists and is continuous.
For all $r\ge 0$ a function $g$ in an open set $C^{r+1}$ iff $g'$ is of class $C^r$, a composite of $C^r$ functions is $C^r$, and the function $t\mapsto 1/t$ is of class $C^r$.
I am attempting to prove this by induction. For my base step a assume $f$ is of class $C^1$. Then by definition of $C^1$ I know that $f'$ exists and is continuous. Is it enough to say that $(f^{-1})'$ exists because $(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}$?
Then I'm not sure where to go with the induction step yet
Recall that if $f:(a,b)\to (c,d)$ is a continuous bijection, then so is $f^{-1}.$ Use this, the nonvanishing of $f',$ and the formula $$\tag 1 (f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$$ to show $f\in C^1 \implies f^{-1}\in C^1.$
For the induction you could argue this way: Suppose $f\in C^r$ and $f^{-1}\in C^k$ for some $k\in \{1,2,\dots ,r-1\}.$ Then $f^{-1}\in C^{k+1}.$ Proof: The right of $(1)$ is the composition of $1/x\in C^\infty$ with $f'\circ f^{-1}.$ Because $f'\in C^{r-1}$ and $f^{-1}\in C^k,$ the compostion is $C^k.$ Thus $(f^{-1})'\in C^k,$ which is the same as saying $f^{-1}\in C^{k+1}.$