It' pretty much as stated in the title. Looking for a proof of the statement that $\mathbf{x}\wedge \mathbf{w} =0$ iff $\mathbf{w}=\mathbf{x}\wedge \mathbf{w}'$ where $\mathbf{x}, \mathbf{w}' \in V$ where $V$ is some $n$-dimensional vector space.
The backwards direction is clear to me, it's the forwards direction that's causing me the issue (although I think I'm just being dense and it's trivial). I have tried picking a basis for $V$, say $ \{ \mathbf{e}_1, \ldots, \mathbf{e}_n \}$ and setting $\mathbf{e}_1=\mathbf{x}$ WLOG but not getting very far.
Assume $x\neq 0$. Then, as you said, we can pick a basis $\{ x,e_{2},...,e_{n} \}$ for $V$. Writing $e_{1}=x$, the set $\{ e_{i}\wedge e_{j}\colon i<j \}$ is a basis for $\Lambda^{2}V$.
Now, write $w=w'x+\sum_{i>1}w_{i}e_{i}$. The exterior product $x\wedge w$ satisfies
$$ x\wedge w=w'(x\wedge x)+\sum_{i=2}^{n}w_{i}x\wedge e_{i}=\sum_{i=2}^{n}w_{i}e_{1}\wedge e_{i}. $$
Since the vectors $e_{1}\wedge e_{2},...,e_{1}\wedge e_{n}$ are independent, we get $x\wedge w=0$ iff $w_{2}=...=w_{n}=0$, which is equivalent to $w=w'x$.
Hope this helps!