Let us consider a finite dimensional vector space $V$ over a field $\mathbb K$ of characteristic zero (we can take $\mathbb K=\mathbb R$ or $\mathbb C$, for example). Let $k$ be the dimension of $V$: when dealing with vectors in $V^{\otimes k}$ or ${V^{*}}^{\otimes k}$ we refer to $k$ as the "weight" of the tensor. This terminology is standard in homological algebra.
To reconstruct the result you want to prove, we need to consider a couple of maps. The first one is the alternating map
$$Alt^{k}_{*}: \wedge^{k}V^{*}\rightarrow Alt^k(V^{*})\subset V^{*}\otimes\dots\otimes V^{*},$$
with
$$Alt^{k}_{*}(\omega_1\wedge\dots\wedge \omega_k):=\frac{1}{k!}\sum_{\pi\in S_k}(-1)^{\pi}
\omega_{\pi(1)}\otimes\dots\otimes\omega_{\pi(k)},$$
denoting by $Alt^{k}(V^{*}):=Im(Alt^{k}_{*})$ the linear subspace of $V^{*}\otimes\dots\otimes V^{*}$ ($k$-times) consisting of all "alternating" tensors of weight $k$. $Alt^{k}_{*}$ is an isomorphism.
The second map we need to consider is
$$\varphi:(V^{*}\otimes\dots\otimes V^{*})\otimes(V\otimes\dots\otimes V)\rightarrow\mathbb K,$$
where
$$\varphi(\omega_{1}\otimes\dots\otimes\omega_{k}\otimes v_{1}\otimes\dots\otimes v_{k}):=\prod_{r=1}^k\omega_r(v_r).$$
$\varphi$ is the (I d not discuss unicity here) multilinear extension of the duality $V^{*}\otimes V\rightarrow \mathbb K$ to the components of weight $k$ of the tensor algebras
$$ T(V^{*})=\bigoplus_{k\geq 0} {V^{*}}^{\otimes k} $$
$$T(V)=\bigoplus_{k\geq 0} V^{\otimes k} $$
In summary, we want to characterize the map (always focusing on weight $k$ tensors)
$$\varphi\circ(Alt^{k}_{*}\otimes 1): \wedge^{k}V^{*}\otimes (V\otimes\dots\otimes V
)\rightarrow \mathbb K.$$
The above map gives us the formula of the "pairing" in your question.
It follows that, choosing a basis $\{e_i\}$ for $V$ and the dual basis $\{\omega_i\}$ for $V^{*}$ the composition
$$\varphi(Alt_{*}(\omega_1\wedge\dots\wedge\omega_k)\otimes(e_1\otimes \dots\otimes e_k))=\frac{1}{k!}\sum_{\pi\in S_k}(-1)^{\pi}\varphi(
\omega_{\pi(1)}(e_1)\otimes\dots\otimes\omega_{\pi(k)}(e_k))=\\
=\frac{1}{k!}\sum_{\pi\in S_k}(-1)^{\pi}\prod_{i=1}^k\omega_{\pi(i)}(e_i)=\frac{1}{k!}\det(A), $$
where $A$ is the $k\otimes k$ matrix with entries $A_{ij}:=\omega_{i}(e_j)$.
Considering dependent vectors $\omega_{*}$ results in having a zero determinant.