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.


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".