Uniqueness of Wedge Product

Question

In Lee's smooth manifolds it is an exercise to show that the wedge product $\wedge \colon \Lambda^k(V)\times \Lambda^l(V) \to \Lambda^{k+l}(V)$ is the unique associative, bilinear and anticommutative map satisfying $$\varepsilon^{i_1} \wedge \cdots \wedge \varepsilon^{i_k} = \varepsilon^I \;\;\;\;\; (\star)$$ To clarify, given a multi-index $I = (i_1,\dots, i_k)$, $\varepsilon^I$ is the covariant $k$-tensor in $\Lambda^k(V)$ defined by $$\varepsilon^I(v_1,\dots,v_k) = \det\begin{pmatrix} \varepsilon^{i_1}(v_1) & \cdots & \varepsilon^{i_1}(v_k)\\ \vdots & \ddots & \vdots \\ \varepsilon^{i_k}(v_1) & \cdots & \varepsilon^{i_k}(v_k) \end{pmatrix}$$ If we take $\omega \in \Lambda^k(V)$ and $\eta \in \Lambda^l(V)$ then we can expand in the usual basis as $\omega = \sum_{I}'\omega_I \varepsilon^I$, $\eta = \sum_J'\eta_J\varepsilon^J$. Taking $*$ to be a map satisfying these four properties, we can write \begin{align*} \omega * \eta &= \left(\sum_{I}'\omega_I \varepsilon^I\right)*\left(\sum_J'\eta_J\varepsilon^J\right)\\ &=\sum_I'\omega_I\left(\varepsilon^I*\left(\sum_J'\eta_J\varepsilon^J\right)\right)\\ &=\sum_I'\omega_I\left(\sum_J'\eta_J\left(\varepsilon^I *\varepsilon^J\right)\right) \end{align*} At this point all we've used is bilinearity. So if we can prove $\varepsilon^I *\varepsilon^J = \varepsilon^I \wedge\varepsilon^J$ then we are done. Using associativity and the property $(\star)$ we find \begin{align*} \varepsilon^I *\varepsilon^J &= (\varepsilon^{i_1} *\cdots *\varepsilon^{i_k})*( \varepsilon^{j_1} *\cdots*\varepsilon^{j_l})\\ &= \varepsilon^{i_1} *\cdots *\varepsilon^{i_k}* \varepsilon^{j_1} *\cdots*\varepsilon^{j_l}\\ &= \varepsilon^{IJ}\\ &= \varepsilon^I \wedge \varepsilon^J \end{align*} Here the first equality is $(\star)$, the second equality is associativity, the third is $(\star)$ again, and the fourth is an earlier proposition in Lee about the wedge product. But this argument does not use anticommutativity anywhere. What am I missing?