Some notation
Let $ V $ be a finite dimensional vector space (say, over the real numbers). Let's suppose that all we know about the exterior power $ \bigwedge^k V^* $ of the dual space $ V^* $ of $ V $ is its universal property.
We can then define a bilinear map $$ \bigwedge\nolimits^k V^*\times \bigwedge\nolimits^l V^*\to \bigwedge\nolimits^{k+l} V^* $$ that sends a wedge $ \alpha_1\wedge\dots\wedge \alpha_k $ and a wedge $ \beta_1\wedge\dots\wedge \beta_l $ in the wedge $ \alpha_1\wedge\dots\wedge \alpha_k\wedge \beta_1\wedge\dots\wedge \beta_l $ and more generally a $ k $-form $ \omega $ and an $ l $-form $ \eta $ into some $ k + l $ form which we denote $ \omega\wedge \eta $.
Now, an $ \omega\in \bigwedge^k V^* $ corresponds to a $ k $-linear alternating map $ \omega\colon V\times\dots\times V\to \mathbb R $ in the following way. It's well know that $$ \bigwedge\nolimits^k V^*\cong \Bigl(\bigwedge\nolimits^k V\Bigr)^* $$ by a map that sends a $ k $-form $ e_{i_1}^*\wedge\dots\wedge e_{i_k}^* $ to $ (e_{i_1}\wedge\dots\wedge e_{i_k})^* $, where $ \{e_1,\dots,e_n\} $ is a basis of $ V $, $ \{e_1^*,\dots,e_n^*\} $ is the dual basis on $ V^* $, and $ \{e_{i_1}\wedge\dots\wedge e_{i_k}\} $ and $ \{(e_{i_1}\wedge\dots\wedge e_{i_k})^*\} $ are respectively the induced basis on $ \bigwedge^k V $ and its dual basis. So given such an $ \omega\in \bigwedge^k V^* $ take $ \tilde\omega\in \bigl(\bigwedge^k V\bigr)^* $ and compose this object with the canonical $ k $-linear map $ V\times\dots\times V\to \bigwedge^k V $.
My question
Suppose that I take a $ k $-form $ \omega $ and an $ l $-form $ \eta $. I then take their wedge product $ \omega\wedge \eta $ as outlined above. And in the end I go to a $ k + l $-linear alternating map (which I give the same name) $ \omega\wedge \eta\colon V\times\dots\times V\to \mathbb R $.
I was wondering if this map $ \omega\wedge \eta\colon V\times\dots\times V\to \mathbb R $ is the same thing as the "wedge product" of the two $ k $- and $ l $-linear maps $ \omega\colon V\times\dots\times V\to \mathbb R $ and $ \eta\colon V\times\dots\times V\to \mathbb R $ as it is taught in differential geometry textbooks.
In other words, define $$ \omega\barwedge \eta = \frac{(k + l)!}{k!l!}\mathrm{Alt}(\omega\otimes \eta) $$ where $ \omega\otimes \eta $ is the tensor product of multilinear maps and $ \mathrm{Alt} $ is the usual alternator operator. Is it true that $$ \omega\wedge \eta = \omega\barwedge \eta\,\text{?} $$