A star shaped open set in $\mathbb{R}^n$ is diffeomorphic to $\mathbb{R}^n$ After googling this fact (which is used in Bott and Tu to show the existence of good covers of manifolds) I've gotten the impression this is somewhat difficult to prove.  But I also came across this homework problem (problem 0) with a hint:  http://www.math.columbia.edu/~thaddeus/geometry/hw10.pdf
which gives me the impression that it is maybe not so difficult as I thought.  If anyone has any thoughts about how to prove this result I'd be very appreciative. Thanks for your time.
 A: I gave an answar here as well.

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)
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 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 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.$\square$
A: Let $U$ be an open  subset of $\mathbb R^n$ that is star-shaped with respect to the origin $0\in U$. Let $f:U\mapsto (0,\infty)$ be a regularized distance function, i.e., a $C^\infty$ function  such that $f(x)/\operatorname{dist}(x,\partial U)$ is pinched between two positive constants. 

Aside: how to construct such $U$? Take all maximal dyadic cubes $Q_k\subset U$ such that $\operatorname{dist} (Q_k, \partial U)\ge \operatorname{diam} Q_k  $ (aka Whitney decomposition of $U$). Let $\varphi_k$ be a smooth  partition of unity associated to the cover of $U$ by larger cubes $\frac32Q_k$. Then $f(x)=\sum \operatorname{dist} (Q_k, \partial U)\,\varphi_k(x)$.  

Considering the construction above, it's clear that we can make $f$ constant in a neighborhood of $0$, say $f=K$ there. Pick $r>0$ small enough so that the sphere $r\,\mathbb S^{n-1}$ is contained in this neighborhood. For each unit vector $\xi\in \mathbb S^{n-1}$, let $\gamma_\xi:\mathbb R\to U$ be the solution of the ODE $\dot \gamma_\xi=f(\gamma)\,\gamma$ with initial value $\gamma_\xi(0)=r\xi$. Observe that $\gamma_\xi(t)=e^{Kt}r \xi$ for $t\le 0$. For $t>0$, the integral curve still goes in the direction $\xi$, but slows down approaching $\partial U$ and never leaves $U$. It should be clear that the integral curves sweep out $U$.
Define a map from $\mathbb R^n$ onto $U$ by 
$$F(\rho\, \xi)= \gamma_\xi( K^{-1}\log \rho  ),\quad \rho> 0, \ \xi\in \mathbb S^{n-1} \tag1$$
On the unit ball $F$ is linear. It is $C^\infty$ smooth everywhere. It is a bijection onto $U$. The invertibility of its derivative follows from ODE theorems on the dependence of  solutions on initial values, see Hartman. 
