Covectors $\omega^1, ..., \omega^k$ are linearly dependent iff their wedge product is zero

Question

How can I prove that covectors $\omega^1, ..., \omega^k$ are linearly independent iff their wedge product $\omega^1\wedge ...\wedge \omega^k$ is not zero?

Answer 1

"If part:" Suppose that $\omega^1\wedge ...\wedge \omega^k$ is not zero. Suppose that $a_i\in\mathbb{R}$ where $1\leq i\leq k$ such that $$\tag{1}a_1\omega^1+\cdots+a_k\omega^k=0,$$ the zero covector. Take the wedge product with $\omega^1\wedge \omega^2\wedge\cdots\wedge \omega^{i-1}\wedge\omega^{i+1}\wedge\cdots\wedge\omega^k$ with $(1)$, we get $$a_i\omega_i\wedge\omega^1\wedge \omega^2\wedge\cdots\wedge \omega^{i-1}\wedge\omega^{i+1}\wedge\cdots\wedge\omega^k=(-1)^{i-1}a_i\omega^1\wedge ...\wedge \omega^k=0,$$ where $1\leq i\leq k$. Therefore, if $\omega^1\wedge ...\wedge \omega^k\neq 0$, we must have $a_i=0$ for $1\leq i\leq k$. By definition, $\omega^1, ..., \omega^k$ are linearly independent.

After typing the "if part", I find that I forgot about the proof of "only if part". Hope that someone can give the full proof.