If we think of the determinant as a multilinear map from the set of $n$-column vectors to $\mathbb{R}$, $$\det:\mathbb{R}^n\times\cdots\times\mathbb{R}^n\to\mathbb{R},$$ am I right in saying that $$\det=e^1\wedge\cdots\wedge e^n,$$ where $\{e^i\}$ is the basis dual to the standard basis of $\mathbb{R}^n$ and $\wedge$ is the wedge product?
I think we can prove it by just expanding both sides, using Leibniz formula for the determinant and the definition $$\omega\wedge\eta=\frac{(k+l)!}{k!l!}\rm{Alt}(\omega\otimes\eta)$$ for the wedge product of $\omega\in\Lambda^k$, $\eta\in\Lambda^l$.
Recall that if $\dim(E) = n$ then $\dim(\Lambda^n(E))$ = 1.
Therefore $\det$ and $e^1 \wedge \dots \wedge e^n$ are colinear. Since they agree (and do non cancel) on the stardard basis of $\Bbb R^n$, they are actually equal.