The volume of a parallelepiped $(V)$ is given by the triple scalar product:
$$V=\mathbf{c}\cdot{}(\mathbf{a}\times\mathbf{b})$$
where $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are the vectors originating at one of the vertices of the parallelepiped.
Is there a way to easily extend this formula to higher dimensions, analogous to the way that the volume of a sphere can be extended to higher dimensions?
In linear algebra it would be useful to have a general formula for the volume of a hyper-parallelepiped because the determinant of a $3 \times 3$ Gram matrix (and metric tensor too, I think) is just $V^2$. In general, the triple scalar product offers a formula for deriving the linearly independent basis vectors of a $3\times3$ matrix in terms of each other and the constant $V$. It would be nice to extend this trick to $n \times n$ matrices.
In $n$ dimensions let $\varepsilon_{i_1\dots i_n}$ be the Levi-Civita Symbol defined by
$$\varepsilon_{i_1\dots i_n}=\begin{cases} 1&\text{ if }i_1,\dots ,i_n\text{ is an even permutaion of }1,\dots ,n\\ -1&\text{ if }i_1,\dots, i_n\text{ is an odd permutaion of }1,\dots ,n\\ 0&\text{ otherwise.}\\ \end{cases} $$
Then the parallelotope given by the vectors $\mathbf{a}^1\dots \mathbf{a}^n$ has content
$$\sum_{i_1,\dots, i_n}\varepsilon_{i_1\dots i_n}{a}^1_{i_1}\dots {a}^n_{i_n}$$
which agrees with the $3$-dimensional case because
$$(\mathbf{a}\times \mathbf{b})\cdot\mathbf{c}=\sum_{i,j,k}\varepsilon_{ijk}a_ib_jc_k$$