In Wikipedia's motivating example of wedge product, what happened to $e_1 \wedge e_1$

Question

Wikipedia (https://en.wikipedia.org/wiki/Exterior_algebra) has a motivating example for wedge product.

In particular, it was written that:

$ \begin{align} {\mathbf v}\wedge {\mathbf w} & = (a{\mathbf e}_1 + b{\mathbf e}_2) \wedge (c{\mathbf e}_1 + d{\mathbf e}_2) \\ & = ac{\mathbf e}_1 \wedge {\mathbf e}_1+ ad{\mathbf e}_1 \wedge {\mathbf e}_2+bc{\mathbf e}_2 \wedge {\mathbf e}_1+bd{\mathbf e}_2 \wedge {\mathbf e}_2 \\ & =(ad-bc){\mathbf e}_1 \wedge {\mathbf e}_2 \end{align} $

It is explained that:

where the first step uses the distributive law for the exterior product, and the last uses the fact that the exterior product is alternating, and in particular $e_2 ∧ e_1 = −e_1 ∧ e_2$.

In this example, whatever happened to the term associated with $e_1 \wedge e_1$? and $e_2 \wedge e_2$?

Wikipedia doesn't say.

Answer 1

We have $$e_1\wedge e_1=-e_1\wedge e_1,$$ so $e_1\wedge e_1=0$.

Answer 2

Since the product is alternating, $$\color{red}{e_1}\wedge\color{blue}{e_1} = -\color{blue}{e_1}\wedge\color{red}{e_1}$$ and therefore $$e_1\wedge e_1 = 0$$

Answer 3

The wedge product is defined to be the asymmetric tensors $$ v_1 \wedge v_2 = v_1 \otimes v_2 - v_2 \otimes v_1 $$ Therefore $e_1 \wedge e_1 = 0$ vanishes.