I'm reading John Browne's Grassmann Algebra, Vol 1 : Foundations. Early on, he asserts without proof that if $x$ and $y$ are any two vectors in the underlying (real) vector space such that $x \wedge y = 0$, then $x$ and $y$ are linearly dependent. Take the vector space to be $R^3$, say. The result is equivalent to proving that if $e_i, e_j$ are two of the standard basis vectors, then $e_i \wedge e_j \neq 0$.
In the framework of axioms and or constructions that Browne provides, how does one prove that simple fact?
The point is that the determinant gives you a non-zero linear function from $\wedge^n V$ to the ground field (which can be arbitrary) when $n=\mathrm{dim}(V)$. So this space is non-zero. Now if you have a basis $e_1,\dots,e_n$ of $V$, then multilinearity and skew-commutativity together imply $e_1 \wedge \cdots \wedge e_n$ spans the top exterior power, and must therefore be non-zero. The result you want follows.
On the other hand, as you point out in the comments below and I confirm, it seems that Browne's axioms are not enough to imply this fact about determinant, and indeed the result you want might fail: any quotient of the Grassmann algebra also satisfies his axioms.