The Clifford product of a pair of vectors $a,b$ is an associative operation defined by
$$ ab = a \cdot b + a \wedge b.$$
In sufficiently low dimensions I am used to being able to define the Clifford product on arbitrary $k$-vectors by repeatedly applying the vector definition. For instance, suppose that I build a Clifford algebra over $\mathbb{R}^3$ with the usual (positive) Euclidean inner product. Then I can easily write out the Clifford product of any pair of basis bivectors. For instance,
$$ e_{12}e_{13} = e_1 e_2 e_1 e_3 = -e_2(e_1 e_1)e_3 = -e_2 (1) e_3 = -e_{23}.$$
In four dimensions I get stuck, because it's possible that two of the indices don't "cancel," and then I don't know how to apply the product:
$$e_{12}e_{34} = e_1 e_2 e_3 e_4.$$
Where do I go from here? I strongly suspect that this equals just $e_{1234}$, but I don't know how to show (in an explicit, pedantic, algebraic way) that
$$e_1 e_2 e_3 e_4 = e_1 \wedge e_2 \wedge e_3 \wedge e_4.$$
Thanks!