Is it true that, for a $k-$form $\alpha$, $\alpha\wedge\alpha$ is always $0$ if $k$ is odd; and not always $0$ if $k$ is even?
My reason is that, for basic $k-$forms $dx_I,dx_J$, we have $dx_J\wedge dx_I=(-1)^{k^2}dx_I\wedge dx_J$, so, if $k$ is odd, they always cancel out each other.
Yes it is always true. The wedge product is indeed a generalization of the vector product, and the vector product of the same vector is easily seen to be always zero. Like the cross product, the exterior product is anticommutative, meaning that $ u \wedge v=-(v \wedge u)$ for all vectors $u$ and $v$.