Suppose $e_1,\ldots,e_n$ and $b_1,\ldots,b_n$ are bases for $E$. Then for all $u\in\Hom(E)$:
$$
\det(b_j^*(u(b_k)))=\det(e_j^*(u(e_k)))
$$
i.e. the determinant of the matrix of $u$ with respect to a basis does not depend on the basis.
Let $\o$ and $\eta$ be volume forms on $E$ such that $\o(e_1,\ldots,e_n)=\eta(b_1,\ldots,b_n)=1$. Then by definition of $\det u$ and its independence of volume forms:
$$
\det(b_j^*(u(b_k)))
=\o(u(b_1),\ldots,u(b_n))
=\det u
=\eta(u(e_1),\ldots,u(e_n))
=\det(e_j^*(u(e_k)))~.
$$