Is it true that a connected, simply connected, nilpotent $n$-dimensional Lie group $G$ is homeomorphic to $\mathbb R^n$?
EDIT: Maybe a possible argument is the following: Since $G$ is simply connected, $G$ cannot contain any non-trivial maximal compact subgroups. By a theorem associated to Iwasawa and Malcev, all maximal compact subgroups are conjugate and thus have the same dimension. By a theorem by Hochschild(?), $G/K$ is diffeomorphic to $\mathbb R^n$, where $K$ is a(ny) maximal compact subgroup. But for $G$ simply connected, connected, nilpotent, $K$ must be trivial, whence $G$ itself is diffeomorphic to $\mathbb R^n$.