I've been away from math far too long, and now my memory plays tricks on me when I try to recall simple facts.
I know that $\det(v_1,\ldots, v_n)$ is the oriented volume of the simplex determined by the origin and the vectors $v_1, \ldots, v_n$ (up to a constant factor depending on the dimension $n$).
But I fell into utter confusion when I tried to make up why, especially when I tried to prove that this formula works for a shifted simplex, too.
What I'm sure is that $\det$ is an alternating bi-linear form that fits naturally with an oriented volume function.
What I've found on the net is dependent on what definition is used, and prone to circular reasoning.
Explicit question: how and why is the oriented volume of a simplex related to the determinant of its vectors?
What I've tried: proved it for $n=1, n=2, n=3$, and I've seen that this is not the way to go.
Here is a simple conceptual way of doing it. The determinant function is invariant under the operation of adding one row to another, it is multiplied by $c$ when multiplying a row by $c$, and changes sign when two rows are exchanged, finally the determinant of the identity is $1$. Now it is easy to show that there is at most one such function with those properties. (Do row reduction.)
Now the volume, suitably modified to be signed has all these properties. The first adding a multiple of one vector to another is a shear transform. Draw a picture of parallelograms, two dimensions suffice cause only two vectors are involved. Multiplying a vector by $c$, also multiples the volume by $c$, taking care of signs. The conclusion is that the two functions are equal.