So not so long ago, I asked for a simple proof that $\mathbf{R}$ has only one smooth structure. A proof that was communicated to me by Ryan Budney (link text) was the following:

So let me recall his argument: So let $X$ be a real line endowed with a "potentially" exotic smooth structure. We know that $X$ is Hausdorff and paracompact so for every open covering $\mathcal{U}$ of $X$ we have a partition of unity dominated by $\mathcal{U}$. Using this we can endow $\mathbf{R}$ with a Riemanian metric $ds^2$ (choose your favorite open covering which is locally finite!). Let $x_0$ be a point of $X$ so that $X-x_0=X^+\bigcup X^{-}$ is the disjoint union of the two components. Finally, note that one may integrate this metric against the Haar measure of $X$ using the velocity vectors $1$ and $-1$ in the fiber above $x_0$ to get two bijections

$f^+:X^+\rightarrow\mathbf{R}_{>0}$

and

$f^-:X^-\rightarrow\mathbf{R}_{<0}$.

Since the metric $ds^2$ is smooth we see that $f^+$ and $f^-$ are smooth and that they glue in a smooth way. So basically, the proof works because we can think of $\mathbf{R}$ as the union of two geodesics.

Q: Is there somekind of similar argument for $\mathbf{R}^2$ and $\mathbf{R}^3$ ?

Any simple proof along different lines is welcome!

e.g.mathoverflow.net/questions/18844;thegeodesic through one of its points is a surjective map $\mathbb R\to M$, according to Hopf-Rinow. The map is locally a diffeo by the inverse function theorem, and if it were not injective it would be periodic, by uniqueness of geodesics---but in that case $M$ would be compact. – Mariano Suárez-Alvarez♦ Mar 14 '11 at 22:06