I'm pretty certain the answer is no provided $n \geq 3$. Let $X$ be the Whitehead manifold: http://en.wikipedia.org/wiki/Whitehead_manifold
It's a contractible open 3-manifold which is not homeomorphic to $\mathbb R^3$.
$X^2$ I claim is homeomorphic to $\mathbb R^6$. I don't have a slick proof of this. The idea is, ask yourself if you can put a boundary on $X^2$ to make $X^2$ the interior of a compact manifold with boundary. Larry Siebenmann's dissertation says if $X^2$ is simply-connected at infinity, you're okay. But the fundamental-group of the end of $X^2$ has a presentation of the form $(\pi_1 X Y * \pi_1 XY) / <\pi_1 X^2>$Y^2>$, where$*$denotes free product and angle brackets "normal closure". This , and$\pi_1 Y$is "the fundamental group at infinity for$X$". So$\pi_1 X^2$is the trivial group. Once you have it as the interior of a compact manifold with boundary, the h-cobordism theorem kicks in and tells you this manifold is$D^6$. 1 I'm pretty certain the answer is no provided$n \geq 3$. Let$X$be the Whitehead manifold: http://en.wikipedia.org/wiki/Whitehead_manifold It's a contractible open 3-manifold which is not homeomorphic to$\mathbb R^3$.$X^2$I claim is homeomorphic to$\mathbb R^6$. I don't have a slick proof of this. The idea is, ask yourself if you can put a boundary on$X^2$to make$X^2$the interior of a compact manifold with boundary. Larry Siebenmann's dissertation says if$X^2$is simply-connected at infinity, you're okay. But the fundamental-group of the end of$X^2$has a presentation of the form$(\pi_1 X * \pi_1 X) / <\pi_1 X^2>$, where$*$denotes free product and angle brackets "normal closure". This is the trivial group. Once you have it as the interior of a compact manifold with boundary, the h-cobordism theorem kicks in and tells you this manifold is$D^6\$.