Show that the magnitude of the exterior product is equal to the area of the parallelogram, and extends to volumes

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Perhaps my Title is not worded well enough, please let me explain my question:

I have read the wonderful answer here on the connection between the exterior product and the cross product. It says

Given two vectors $\vec v,\vec w\in \Bbb L^n$, we define the bivector $\vec v \wedge \vec w\in \Lambda \Bbb L^n$ as the oriented plane segment whose area(/norm) is equal to the area of the parallelogram with sides $\vec v$ and $\vec w$

I am confused with how this generalizes. For example, why does it follow that if I take an exterior product of $3$ vectors in $\mathbb{R}^n$, it again hold that the length of the resulting trivector is the volume of the paralelepiped they form. All this seems very elegant to me, a very nice generalization of vectors, and I want to understand how it all comes toghether.

I would be thankful to anyone who can offer a few words on how this follows. I thank you in advance