Let $e_{1},\ldots,e_{n}$ be vectors in $\mathbb{R}^{n}$. Define a parallelepiped $P$ to be a translate of the set
$$\left\{x\in\mathbb{R}^{n} : x=t^{1}e_{1}+\cdots+t^{n}e_{n}, 0\leq t^{i}\leq 1\right\}$$
Define the volume of $P$ by $\text{vol}(P):=\left|\det(A)\right|$, where $A$ is the matrix with columns $e_{1},\ldots,e_{n}$.
How does one proof directly from this definition of the volume of parallelepiped that if $I_{1},\ldots,I_{n}$ are nonoverlapping (i.e. disjoint interiors) parallelepipeds contained in $P$, then
$$\sum_{j=1}^{n}\text{vol}(I_{j})\leq\text{vol}(P)$$
This inequality seems geometrically evident from "cutting up" $P$, but I am having trouble making this rigorous. It seems like the easiest way to show this is to prove that this definition can be realized by Riemann integration of indicator functions using the change of variable formula and then use the additivity properties of the integral. But I want to avoid this.
As a simple case, how would one show that a parallelepiped (not necessarily of the same orientation) $I\subset P$ satisfies $\text{vol}(I)\leq\text{vol}(P)$ directly?