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I've seen the following claim several times:

If $V$ is a vector space over $K$ with basis $\{e_1,\ldots,e_n\}$ then the basis to the kth exterior power of $V$ is given by the elements $$\{e_{i_1}\wedge e_{i_2}\wedge\cdots\wedge e_{i_k} \mid 1 \le i_1 < i_2 < \cdots < i_k \le n\}$$

Now, given some basic facts on exterior powers, it's not hard to show that the above set actually spans the $k$th exterior power. On the other hand, I've never seen a complete proof of the linear independence of this set. So my question is: What is the simplest way to show that $\{e_{i_1}\wedge e_{i_2}\wedge\cdots \wedge e_{i_k} \mid 1 \le i_1 < i_2 < \cdots < i_k \le n\}$ is linearly independent?

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Careful; you use $k$ for both the underlying field and for an integer. –  Qiaochu Yuan Jul 24 '11 at 23:29
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2 Answers

We induct on $n - k$.

Suppose there exists a nontrivial linear dependence among the pure tensors you list in $\Lambda^k(V)$. Then not all of the tensors share the same components $e_i$, so there is some $i$ such that some tensor does not contain $e_i$ (and some other tensor does). Take the exterior product with $e_i$; this gives a linear dependence among a smaller number of pure tensors in $\Lambda^{k+1}(V)$. By repeating this argument we see that it suffices to show that no single pure tensor can be equal to zero. If $k < n$, then it suffices to take the exterior product with the rest of the basis so that we reduce to the case $k = n$. If $k = n$, it suffices to use any of the standard proofs that the determinant exists.

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If you want an easy access reference, this book, which is available for free online has a proof of the statement you're looking for in section 2.3.2, basically it's their "lemma 3".

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