# Question about higher differentiability of inverse function

Assume that $f:I: \rightarrow J$ is a bijection of class $C^1$ and $f'(x)>0$ for all $x \in I$, where $I$, $J$ are intervals in $\mathbb R$. Then by the known theorem $f^{-1}$ is of class $C^1$ and $(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}$ for $y\in J$.

Assume now that $f$ is of class $C^n$, $n \in \mathbb N$. Why $f^{-1}$ is also of class $C^n$?

-
It seems to me that proof is by induction. If $f^{-1}$ is of class $C^n$ for some $n$. Then by formula on $(f^{-1})'$ we have that $f'\circ f^{-1}$ is of class $C^n$ whence $f^{-1}$ is of class $C^{n+1}$. –  L.T Oct 16 '12 at 16:13