You've just given a well-known construct in exterior algebra. (The analogy with the exterior derivative is really not right, because this construction will generalize to something linear over the $C^\infty$ functions, namely wedging with a fixed $1$-form on the manifold, whereas $d(f\omega) \ne f d\omega$ for general functions $f$.) The appearance of Lie brackets in the exterior derivative formula is, on the other hand, what's needed to make the resulting object a tensor field, namely (multi-)linear over $C^\infty$ functions in the sense that $d\omega(X_0,\dots,fX_i,\dots,X_k) = f d\omega(X_0,\dots,X_i,\dots,X_k)$. There is no such thing when we work in a fixed vector space, as scalars always pull out.
Given any vector $\alpha\in V^*$ and any vector $v\in V$, you can define two different linear maps:
\begin{align*}
e_\alpha\colon& \Lambda^k V^* \to \Lambda^{k+1} V^*, \quad\text{given by } e_\alpha(\tau) = \alpha\wedge\tau \\
\iota_v\colon& \Lambda^k V^* \to \Lambda^{k-1}V^*, \quad\text{given by } \iota_v(\tau)(v_1,\dots,v_{k-1}) = \tau(v,v_1,\dots,v_{k-1}).
\end{align*}
These are often called, respectively, exterior and interior product. Of course, you can write out the definition of the exterior product in terms of the value on $k+1$ vectors, with appropriate constants and signs on permutations. This is part of the standard definition in exterior algebra (if you think of it as a subspace of $\otimes^\bullet V^*$, rather than as a quotient space).
A further tangential comment: If you choose a basis $\{e_1,\dots,e_n\}$ for $V$ and the corresponding dual basis $\{\alpha_1,\dots,\alpha_n\}$ for $V^*$, then $$\iota_{v_i}\big(e_{\alpha_i} \tau\big) = \tau \qquad\text{and}\qquad e_{\alpha_i}\big(\iota_{v_i}\tau\big) = \tau.$$
(Depending on numeric conventions, perhaps some constant may appear.)