Watching Frederich Shullers "Lectures on the Geometric Anatomy of Physics" series, he defines the determinant of an Endomorphism $\phi$ as
$$\det \phi = \frac{w(\phi(e_1),\ldots \phi(e_n))}{w(e_1, \ldots e_n)}$$
where $w$ is the volume form on some n dim vector space V
I've been trying to prove the property that $$\det(\phi \odot \psi) = \det(\phi)\det(\psi)$$ but have had trouble doing so via this definition. It seems the anti-symmetry of $w$ is key, but I can't figure out how to work it in.
Indeed, you can avoid wedge products. But the issue is elsewhere.
The author of the video sweeps some dust under the carpet.
The definition given of the determinant of an endomorphism $\phi$ is in fact the consequence of the following result:
$$\text{for any independent system} \ v_1 \cdots v_n, \ \frac{\det(\phi(v_1), \cdots \phi(v_n))}{\det(v_1 \cdots v_n)} \ \text{is a constant}$$
and to this constant, we give the name $\det(\phi)$. $\square$
The first issue is therefore to be able to establish this result...
Then, we get almost immediately:
$$\underbrace{\frac{\det(\psi \circ \phi(v_1), \cdots \psi \circ \phi(v_n))}{\det(v_1 \cdots v_n)}}_{\det(\psi \circ \phi)}=\underbrace{\frac{\det(\psi \circ \phi(v_1), \cdots \psi \circ \phi(v_n))}{\det(\phi(v_1), \cdots \phi(v_n))}}_{\det(\psi)}\underbrace{\frac{\det(\phi(v_1), \cdots \phi(v_n))}{\det(v_1 \cdots v_n)}}_{\det(\phi)}$$
