Spivak phrases it well (theorem 4.6, page 82):
If \(V\) has dimension \(n\), it follows that
\(\Lambda^n(V)\) has dimension 1. Thus all alternating
\(n\)-tensors on \(V\) are multiples of any non-zero one.
Since the determinant is an example of such a member of
\(\Lambda^n(V)\) it is not surprising to find it in the following
theorem:
Let \(v_1,\ldots,v_n\) be a basis for \(V\) and
let \(\omega\in\Lambda^n(V)\). If \(w_i=\sum_{j=1}^n
a_{ij}v_j\) then
$$
\omega\left(w_1,\ldots,w_n\right)=\det\left(a_{ij}\right)\cdot\omega\left(v_1,\ldots
v_n\right)$$
(see the examples for numerical verification of this).
Neither the zero \(k\)-form, nor scalars, are considered to be a
volume element.