How should I approach a proof like the following for the interior product:
Let $K$ be a field and $V$ a $K$-vector space. And $\omega_i \in V^*$ with $V^*$ the dual space to $V$. It is given that $$\iota_v(\omega_1 \otimes \cdots \otimes \omega_p) = \omega_1(v)\omega_2 \otimes \cdots \otimes \omega_p.$$ Now it is to be proven that: $$\iota_v(\omega_1 \wedge \cdots \wedge \omega_p)=\sum_{i=1}^p (-1)^{i-1}(\iota_v\omega_i) \omega_1\wedge \cdots \wedge \widehat{\omega_i} \wedge \cdots \wedge \omega_p. \tag{1}$$
I tried using the definition of the wedge product through the sum over the permutations of $\operatorname{sgn}(\sigma)$ and the tensor product of the given dual vectors and used the iota function on this definition, because it is given how the tensor product behaves under it. So $$\iota_v(\omega_1 \wedge \cdots \wedge \omega_p) = \iota_v \bigg(\sum_{\sigma\in S_p} \operatorname{sgn}(\sigma) \omega_{\sigma(1)} \otimes \cdots \otimes \omega_{\sigma(p)} \bigg).$$
But wasn't sure whether or not I was allowed then to change the sum over the permutations to a sum over $1$ to $p$.
OK, you've made some serious progress. Now here's a hint to proceed to the next step. Consider dividing $S_p$ up into subsets $S_p(i) = \{\sigma: \sigma(1)=i\}$. You should convince yourself that these are all (right) cosets of $S_{p-1}$. This breaks your sum up into $\sum_{i=1}^p$. Can you finish now?