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This question may be too basic and even silly, but I am new to exterior algebra and reading Wikipedia.

Given $e_1, e_2,\cdots, e_n$ is a standard basis for a vector space $V$, what does $e_1\wedge e_2\cdots \wedge e_n$ actually look like? It is said to be a basis. So, it is a set of vectors? What is its dimension? Is it just a constant with dimension $1$?

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$e_1 \wedge e_2 \wedge \dots \wedge e_n$ is a single vector. The $n$-th exterior power of an $n$-dimensional space, $\bigwedge^n V$, is one dimensional and $e_1 \wedge e_2 \wedge \dots \wedge e_n$ is the single vector you need to span it.

The exterior powers of a vector space are closely related to the subspaces of that vector space (this is called the Grassmann-Plücker embedding). This is because the wedge product has the convenient property that $v_1,\dots,v_k$ are linearly dependent if and only if $v_1 \wedge \dots \wedge v_k = 0$. Thus the non-zero wedges correspond to linearly independent sets of vectors.

If $v_1 \wedge \dots \wedge v_k \ne 0$ then $v_1,\dots,v_k$ are linearly independent and the wedge $v_1 \wedge \dots \wedge v_k$ corresponds to the signed basis $v_1,\dots,v_k$ of the subspace $W = \operatorname{span}(v_1,\dots,v_k)$. You can picture a signed basis as an oriented parallelepiped. For example, with two vectors, $v, w$, the wedge product $v \wedge w$ is the oriented parallelogram with vertices $v, w, v + w$ and the orientation is from $v$ to $w$. There are pictures of this on the Wikipedia entry.

If $V$ is $n$-dimensional, then $V$ has only one $n$-dimensional subspace, namely $V$, which is why $\bigwedge^nV$ is one-dimensional.

In $\mathbf R^n$ we have a notion of an orthogonal complement of a subspace. This gives a natural pairing between a $k$-dimensional subspace $W$ of $\mathbf R^n$ and the $n - k$ dimensional orthogonal subspace, $W^\perp$. Correspondingly, there should be a natural isomorphism

$$ \bigwedge^k \mathbf R^n \longleftrightarrow \bigwedge^{n - k} \mathbf R^n. $$

Given the basis $\{e_{i_1} \wedge \dots \wedge e_{i_k} : 1 \le i_1 \le \dots \le i_k \le n\}$ of $\bigwedge^k \mathbf R^n$, this bijection is described by

$$ e_{i_1} \wedge \dots \wedge e_{i_k} \mapsto e_{j_1} \wedge \dots \wedge e_{j_{n-k}} $$

where $1 \le j_1 \le \dots \le j_{n_k} \le n$ and $\{j_1,\dots,j_{n-k}\} = \{1,\dots,n\} \setminus \{i_1,\dots,i_k\}$. That is, the subspace spanned by a subset $S$ of $\{e_1,\dots,e_n\}$ is mapped to the subspace spanned by the complement, $S^c$. As you can see, this isomorphism makes sense for any $n$-dimensional vector space, not just those with a notion of orthogonal complement.

As an example, for $\mathbf R^3$ we have an isomorphism $\bigwedge^1 \mathbf R^3 = \mathbf R^3$ onto $\bigwedge^2 \mathbf R^3$. If $\hat\imath, \hat\jmath, \hat k$ are the standard basis vectors for $\mathbf R^3$ then the isomorphism is given by

$$ \hat\imath \leftrightarrow \hat\jmath \wedge \hat k,\qquad \hat\jmath \leftrightarrow \hat\imath \wedge \hat k,\qquad \hat k \leftrightarrow \hat\imath \wedge \hat\jmath. $$

Hopefully you see the relation here to the cross product:

$$ \hat\imath = \hat\jmath \times \hat k,\qquad \hat\jmath = \hat\imath \times \hat k,\qquad \hat k = \hat\imath \times \hat\jmath. $$

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