Let $M$ and $N$, with $M$ compact, a couple of topological (differential) manifolds.
What is the easiest/fastest/most elementary way to show that $M\times N$ cannot be homeomorphic (diffeomorphic) to the standard ${\mathbb R}^n$ ?



You can avoid the Kunneth formula, "all" you need is the existence of a single nontrivial homology group of $M$ in positive dimensions, namely $H_{\text{dim} \\, M}(M;\mathbb{Z}/2\mathbb{Z}) = \mathbb{Z}/2\mathbb{Z}$. Once you know that, it follows that the identity map on $M$ is not homotopic to a constant, and so the inclusion map $M \to M \times N$ is not homotopic to a constant. 


$H^{\dim M}((M\times N),\mathbb Z/2\mathbb Z)\ne 0$ (by virtue of Kunneth's formula), which is not the case for $\mathbb R^n$ if $\dim M>0$. 

