If $f$ is smooth and bijective, is $f^{-1}$ smooth, too? Let $U \subseteq \mathbb{R}^m$ be open. Assume that the function $f \colon U \rightarrow f(U)$ is $\mathcal{C}^{\infty}$ and a bijection and its differential $\text{d}f(x)$ is injective for every $x \in U$. Furthermore, assume that $f$ and $f^{-1}$ are continuous. 
Is this enough to conclude that also $f^{-1}$ is $\mathcal{C}^{\infty}$?
I tried to use the inverse function theorem, but this only gives me a locally result and no inverse function $f^{-1}$ with maximal domain $f(U)$.

Write $f^{-1}=:g$. The inverse function theorem contains the formula
$$dg=\iota\circ df\circ g\ ,\tag{1}$$
whereby $\iota$ denotes taking the inverse of a regular linear map $L:\>{\mathbb R}^n\to{\mathbb R}^n$. As $f$ and $\iota$ are $C^\infty$ formula $(1)$ can be used to set up an induction proof that $g$ is $C^\infty$ as well, starting from $g\in C^0$ (which should be included in the proof of the inverse function theorem), and making heavy use of the chain rule.
