Below is a proof found in Gilbert Strang's book and here Why is the determinant the volume of a parallelepiped in any dimensions? that the determinant equals the volume of a box :
To find the volume of a box whose edges are given by a set of vectors $\{v_{1},\ldots,v_{n}\}$ we apply Gram-Schmidt orthogonalization to $\{v_{1},\ldots,v_{n}\}$, so that \begin{eqnarray*} v_{1} & = & v_{1}\\ v_{2} & = & c_{12}v_{1}+v_{2}^{\perp}\\ v_{3} & = & c_{13}v_{1}+c_{23}v_{2}+v_{3}^{\perp}\\ & \vdots \end{eqnarray*} where $v_{2}^{\perp}$ is orthogonal to $v_{1}$; and $v_{3}^{\perp}$ is orthogonal to $span\left\{ v_{1},v_{2}\right\} $, etc. Since determinant is multilinear, anti-symmetric, then \begin{eqnarray*} \det\left(v_{1},v_{2},v_{3},\ldots,v_{n}\right) & = & \det\left(v_{1},c_{12}v_{1}+v_{2}^{\perp},c_{13}v_{1}+c_{23}v_{2}+v_{3}^{\perp},\ldots\right)\\ & = & \det\left(v_{1},v_{2}^{\perp},v_{3}^{\perp},\ldots,v_{n}^{\perp}\right)\\ & =?& \mbox{signed volume}\left(v_{1},\ldots,v_{n}\right) \end{eqnarray*}
Question: This only proves that the determinant of the original edges equals the volume of the new created box.
That is we don't show $\det(v_1,...,v_n)$ = signed volume $(v_1,...v_n)$. What we do prove is $\det(v_1,...,v_n)$ = Volume of the box with orthogonal edges.
A sketch: Consider a parallelepided spanned by vectors $(v_1,v_2,...,v_n)$: $$ K(v_1,v_2,...,v_n)= \{ \sum_{k=1}^n t_k v_k : 0 \leq t_k\leq 1, 1\leq k\leq n\}$$
Given $c\in [0,1]$ consider an elementary shear of this solid in the 1-2 direction: $K'=K(v_1,v_2-cv_1,v_3,...,v_n)$. Note that the directions $v_3,...,v_n$ are left untouched. $K'$ may be cut into two pieces $K'=(K'\cap K) \cup (K'\setminus K)=: K_1\cup K_2$ (make a drawing). If you translate $K_2$ by the vector $-v_1$ you almost get $K\setminus K'$. True for the interior but not for the borders which, however, are of zero volume. $K$ is in this way reconstructed from $K'$ by cutting the latter into two pieces and translating one of them (up to zero volume corrections). Thus, the volume of $K$ and $K'$ are the same and as you see by inspection, the determinants as well. Repeating the operation and also using the inverse you see that the same holds true for all $c\in {\Bbb R}$.
The transformation described by Strang may be decomposed into $n(n-1)/2$ such operations.