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I believe this theorem that you mentioned only works for 2-vectors. If $w$ is a 2-vector then $w$ has rank $p$, if $\wedge^pw\neq 0$ and $\wedge^{p+1}w= 0$. Actually, let us prove that $w=e_0\wedge(e_1\wedge e_2+e_3\wedge e_4)$ is not decomposable, if $e_0,e_1,e_2,e_3,e_4$ are linear independent vectors. First, $v_1,\ldots,v_n$ are linear independent ...