I sent an e-mail to Erwann Aubry (see Georges Elencwajg's answer) and asked for a scan of this proof from the book "Calcul Différentiel". He replied and send me the proof translated into English. [Here][1] is his original paper. And this is how the proof goes:

> Theorem. Every open star-shaped set $\Omega$ in $\mathbb{R}^n$ is $C^\infty$-diffeomorphic to $\mathbb{R}^n.$ 

Proof. For convenience assume that $\Omega$ is star-shaped at $0.$

Let $F=\mathbb{R}^n\setminus\Omega$ and $\phi:\mathbb{R}^n\rightarrow\mathbb{R}_+$ (here $\mathbb{R}_+=[0,\infty)$) be a $C^\infty$-function such that $F=\phi^{-1}(\{0\}).$ (such $\phi$ exists due to [Whitney extension theorem][2])

Now we set $f:\Omega\rightarrow\mathbb{R}^n$ by formula:
$$f(x)=\overbrace{\left[1+\left(\int_0^1\frac{dv}{\phi(vx)}\right)^2||x||^2\right]}^{\lambda(x)}\cdot x=\left[1+\left(\int_0^{||x||}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot x.$$
Clearly $f$ is smooth on $\Omega.$

We set $A(x)=\sup\{t>0:t\frac{x}{||x||}\in\Omega\}.$ $f$ sends injectively the segment (or ray) $[0,A(x))\frac{x}{||x||}$ to the ray $\mathbb{R_+}\frac{x}{||x||}.$ Moreover $f(0\frac{x}{||X||})=0$ and
$$\lim_{r\rightarrow A(x)}||f(r\frac{x}{||x||})||=\lim_{r\rightarrow A(x)}\left[1+\left(\int_0^{r}\frac{dt}{\phi\left(t\cdot\frac{rx}{||x||}\cdot||\frac{||x||}{rx}||\right)}\right)^2\right]\cdot r=\\ \left[1+\left(\int_0^{A(x)}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot A(x)=+\infty.$$
Indeed, if $A(x)=+\infty,$ then it holds for obvious reason. If $A(x)<+\infty,$ then by definitions of $\phi$ and $A(x)$ we get that $\phi(A(x)\frac{x}{||x||})=0.$ Hence by [Mean value theorem][3] and the fact that $\phi$ is $C^1$
$$\phi\left(r\frac{x}{||x||}\right)\leqslant M(A(x)-r)$$
for some constant $M$ and every $r.$ As a result
$$\int_0^{A(x)}\frac{dt}{\phi\left(t\frac{x}{||x||}\right)}$$
diverges. Hence we infer that $f([0,A(x))\frac{x}{||x||})=\mathbb{R_+}\frac{x}{||x||}$ and so $f(\Omega)=\mathbb{R}^n.$

To end the proof we need to show that $f$ has $C^\infty$-inverse. But as corollary from the [Inverse function theorem][4] we get that it is sufficient to show that $df$ vanish nowhere. 

Suppose that $d_xf(h)=0$ for some $x\in\Omega$ and $h\neq 0.$ From definition of $f$ we get that
$$d_xf(h)=\lambda(x)h+d_x\lambda(h)x.$$
Hence $h=\mu x$ for some $\mu\neq 0$ and from that $x\neq 0.$ As a result $\lambda(x)+d_x\lambda(x)=0.$ But we have that $\lambda(x)\geqslant 1$ and function $g(t):=\lambda(tx)$ is increasing, so $g'(1)=d_x\lambda(x)>0,$ which gives a contradiction.$\square$



  [1]: http://docdro.id/QUWTOmC
  [2]: https://en.wikipedia.org/wiki/Whitney_extension_theorem
  [3]: https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorem_for_vector-valued_functions
  [4]: https://en.wikipedia.org/wiki/Inverse_function_theorem