Definition
If $x_0,...,x_n$ are $(n+1)$ affinely indipendent point of $\Bbb R^n$ (which means that the vectors $(x_1-x_0),...,(x_n-x_0)$ are linearly independent) then simplex determined by them is the set $$ S:=\Biggl\{x\in\Bbb R^n: x=\alpha_1v_1+...+\alpha_nv_n, \sum_{i=1}^n\alpha_i\le1\,\,\,\text{and}\,\,\,\alpha_i\ge0\,\,\,\text{for all}\,i\Biggl\} $$ where $v_i:=(x_i-x_0)$ for each $i>0$.
Definition
If $x_0,...,x_n$ are $(n+1)$ affinely indipendent point of $\Bbb R^n$ then the parallelotopoe determined by them is the set $$ P:=\Big\{x\in\Bbb R^n:x:=\alpha_1v_1+...+\alpha_nv_n,\,\alpha_i\in[0,1]\,\text{for each}\,i\Big\} $$ where $v_i:=(x_i-x_0)$ for each $i>0$.
Now if $n=2$ the simplexes and the parallelotopes are respectively triangles and parallelograms. So we observe that if the point $x,\,y$ and $z$ determine a parallelogram then the simplex $S$ generated by $x,\,y$ and $z$ and the simplex $S'$ generated by $y':=y\,z':=z$ and $t':=x+v+w$ where $v:=(y-x)$ and $w:=(z-x)$ determine a partition of the parallelotope generated by $x,\,y$ and $z$ that is $$ S\cup S'=P\,\,\,\text{and}\,\,\,S\cap S'=\partial S\cap\partial S' $$ as we know by the elementary geometry.
So I ask how to generalise this result: in particular if the volume of a simplex and of a parallelotope are respectively given by the formulas $$ v(S)=\Big|\frac{1}{n!}\det\big[(x_1-x_0),...,(x_n-x_0)\big]\Big|\,\,\,\text{and}\,\,\,v(P)=\Big|\det\big[(x_1-x_0),...,(x_n-x_0)\big]\Big| $$ then I imagine that for any parallelotope there exist $n!$ simplexes whose interiors are disjoint and whose union is the parallelotope. Otherwise if the statement is generally false I ask if there exist $n$ simplexes of volume $\frac 1 n$ whose interiors are disjoint and whose unione is the parallelotope. However I suspect that the proof could be done by induction.
So could someone help me, please?
