Axiomatic explanation of why the volume of a parallelepiped is equal to the area of its base times its height Let $V_n$ be the volume on the set of polytopes in $\mathbb R^n$, defined axiomatically, i.e. a functional, that assigns to each polytope $P\subseteq\mathbb R^n$ a real number $V_n(P)\ge 0$ in such a way that the following conditions are fulfilled: 


*

*For the unit hypercube $C\subseteq\mathbb R^n$ 
$$
V_n(C)=1,
$$

*If $P\cap Q=\varnothing$, then 
$$
V_n(P\cup Q)=V_n(P)+V_n(Q),
$$

*If $P$ is made from $Q$ by a motion, then
$$
V_n(P)=V_n(Q).
$$
Question:


Is there a simple way to prove that the $n$-th volume of a parallelepiped $P_{a_1,...,a_n}$ spanned by $n$ vectors $a_1,...,a_n$ is the poduct of the $n-1$-th volume of $P_{a_1,...,a_{n-1}}$ and the height of $P_{a_1,...,a_n}$ with respect to the base $P_{a_1,...,a_{n-1}}$,
  $$
V_n(P_{a_1,...,a_n})=V_{n-1}(P_{a_1,...,a_{n-1}})\cdot \left|\text{pr}_{\{a_1,...,a_{n-1}\}^\perp}(a_n)\right|
$$
  ?

P.S. I asked this also in MO.
 A: Let $P=P_{a_1,a_2,\dots,a_n}\subset\mathbb{R}^n$ be a parallelopiped with edge vectors $a_1,a_2,\dots, a_n$.  I claim that $P$ is scissors-congruent to $P'=P_{a_1+ra_2,a_2\dots,a_n}$ for any $|r|\leq 1$.  To show this, think of $\mathbb{R}^n$ as having $\{a_1,a_2,\dots,a_n\}$ as its standard basis (it will not matter that this basis is not orthonormal because we will only be using translations).  In this basis, $P=[0,1]^n$ (up to a translation), and we wish to transform it into $P'=Q\times[0,1]^{n-2}$, where $Q$ is the parallelogram with edges $(1,r)$ and $(0,1)$.  This $Q$ is scissors-congruent to $[0,1]^2$ by chopping off a triangle and translating it (as in this diagram).  Crossing this cutting and translation with $[0,1]^{n-2}$, we obtain that $P$ and $P'$ are scissors-congruent.
Now by iterating transformations of this kind (for different orderings of the vectors), we can transform $P$ into a parallelopiped with the same base $P_{a_1,\dots,a_{n-1}}$ but for which $a_n$ has been replaced with its projection onto the orthogonal complement of the base.  For parallelopipeds of this form, it is clear that $V_n(P)$ is proportional to $\|a_n\|$.  Furthermore, if we consider such parallelopipeds with $\|a_n\|=1$, allow the base to vary within the orthogonal complement of $a_n$, and write $W_{n-1}(P_{a_1,\dots,a_{n-1}})=V_n(P)$, $W_{n-1}$ satisfies the axioms for $V_{n-1}$.  By induction, we know that the axioms uniquely determine $V_{n-1}$ (at least on parallelopipeds), so $W_{n-1}=V_{n-1}$.  The desired formula for $V_n(P)$ now follows.
