Two of the properties of the exterior product are the following:
- Let $\psi_1, \ldots , \psi_k, n_{1}, \ldots , n_{\ell}\in V^{\star}$ then it holds that $$\left (\psi_1\land \ldots \land \psi_k\right )\land \left (n_1\land \ldots \land n_{\ell}\right )=\psi_1\land \ldots \land \psi_k\land n_{1}\land \ldots \land n_{\ell}$$
- Let $\omega \in \land^kV^{\star}$ and $\sigma\in \land^{\ell}V^{\star}$. Then it holds that $$\omega\land \sigma=(-1)^{k\ell}\sigma\land \omega$$
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I want to show the second property using besides other the first property.
I have done the following:
Let $\psi_1, \ldots , \psi_k, n_{k+1}, \ldots , n_{k+\ell}\in V^{\star}$. Let $\omega \in \land^kV^{\star}$ and $\sigma\in \land^{\ell}V^{\star}$. Then we have that $\omega= \psi_1\land \ldots \land \psi_k$ and $\sigma=n_{k+1}\land \ldots \land n_{k+\ell}$.
We get the following: \begin{align*}\omega\land \sigma=\left (\psi_1\land \ldots \land \psi_k\right )\land \left (n_{k+1}\land \ldots \land n_{k+\ell}\right ) \\ \overset{ \text{First property }}{ = } \psi_1\land \ldots \land \psi_k\land n_{k+1}\land \ldots \land n_{k+\ell}\end{align*}
Then we have to take a permutation of the form $\pi: (1, \ldots , k, k+1, \ldots , k+\ell) \mapsto (k+1, \ldots , k+\ell, 1, \ldots , k)$, right?
Which is the sign of that permutation?
Does it holds then that \begin{align*}\psi_1\land \ldots \land \psi_k\land n_1\land \ldots \land n_{\ell}&=n_{\pi(k+1)}\land \ldots \land n_{\pi(k+\ell)}\land \psi_{\pi(1)}\land \ldots \land \psi_{\pi(\ell)} \\ & =\text{sign}(\pi)\cdot n_{k+1}\land \ldots \land n_{k}\land \psi_{k+1}\land \ldots \land \psi_{k+\ell} \\ & = \text{sign}(\pi)\left (n_{k+1}\land \ldots \land n_{k}\right )\land \left (\psi_{k+1}\land \ldots \land \psi_{k+\ell}\right )\\ & = \text{sign}(\pi)\sigma\land \omega \end{align*} ?
This answer just records the key points of a chat room discussion about the problem with OP. $\DeclareMathOperator{sgn}{sgn}$
First, for $\omega \in \bigwedge^k V^*$, $\sigma \in \bigwedge^{\ell} V^*$ the anticommutativity identity $$\omega \wedge \sigma = (-1)^{kl} \sigma \wedge \omega$$ cannot be deduced only using the identity $$(\omega_1 \wedge \cdots \wedge \omega_k) \wedge (\sigma_1 \wedge \cdots \wedge \sigma_{\ell}) = \omega_1 \wedge \cdots \wedge \omega_k \wedge \sigma_1 \wedge \cdots \wedge \sigma_{\ell} :$$ After all, that second identity is just a way of recording that the wedge product is associative, and there are perfect good associative operations on the exterior algebra $\bigwedge^{\bullet} V^*$ that are associative but do not satisfy the anticommutativity identity.
So, we need an additional property of $\wedge$. One natural, sufficient property is the one Lord Shark the Unknown suggests, the anticommutativity identity for $1$-forms, i.e., the special case $k = \ell = 1$ of the desired identity: $$\phantom{(\ast) \qquad} \alpha \wedge \beta = - \beta \wedge \alpha \qquad (\ast)$$ for $\alpha, \beta \in \bigwedge^1 V^* \cong V^*$.
Now the claim is almost immediate: One can transpose $\omega_k$ with $\sigma_1, \ldots, \sigma_{\ell}$ in that order, and then the same with $\omega_{k - 1}, \ldots, \omega_1$. In the end we have transposed each of the $k$ $1$-forms $\omega_a$ once with each of the $\ell$ $1$-forms $\sigma_b$, for a total of $k \ell$ transpositions. By the previous paragraph, each of transposition changes the sign of the wedge product, so \begin{align*} \omega \wedge \sigma &= \omega_1 \wedge \cdots \wedge \omega_k \wedge \sigma_1 \wedge \cdots \wedge \sigma_{\ell} \\ &= (-1)^{kl} \sigma_1 \wedge \cdots \wedge \sigma_{\ell} \wedge \omega_1 \wedge \cdots \wedge \omega_k \\ &= (-1)^{kl} \sigma \wedge \omega . \end{align*}