Definition
If $x_0,…,x_n$ are $(n+1)$ affinely independent 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 $i>0$.
So with the previous definition I try to show that the volume of a symplex $S$ is given by the formula
$$
v(S)=\Big|\frac{1}{n!}\det\big[(x_1-x_0),…,(x_n-x_0)\big]\Big|
$$
for each $n\in\Bbb N$.
First of all we observe that any simplex is the intersection of two parallelopides (see here for details) so that any simplex is rectifiable (indeed any parallelopiped is rectifiable and the intersection of rectifiable sets is rectifiable too) and thus any continuous function is there integrable. Now if $x_0,…,x_n$ are $(n+1)$ point affinely independent we define the transformation $h:\Bbb R^n\rightarrow\Bbb R^n$ through the condition
$$
h(x):=A\cdot x+x_0
$$
where $A$ is the matrix whose $j$-th column is the vectors $(x_j-x_0)$ for each $j=1,…,n$. So now we observe that the transformation $h$ carries the simplex
$$
E:=\{x\in\Bbb R^n:x_1+…+x_n\le 1\,\,\,\text{and}\,\,\, x_i\ge 0\,\,\,\text{for all}\,i\}
$$
onto the simplex $S$ generated by the points $x_0,…,x_n$. So if we prove that
$$
v(E):=\frac 1{n!}
$$
for all $n\in\Bbb N$ then by the change variable theorem (it is easy to verify that $h$ is a diffeomorphism)
$$
v(S)=\int_S 1=\int_E|\det A|=\frac 1{n!}|\det A|
$$
for each $n\in\Bbb N$.
So let's start to prove by induction that
$$
v(E)=\frac 1{n!}
$$
for each $n\in\Bbb N$.
So if $n=1$ then $E=[0,1]$ and so clearly the formula trivially holds. So we suppose that the formula holds for $(n-1)$ and we prove that it holds for $n$.
So we have to prove that
$$
\int_E1=\frac 1{n!}
$$
and to do this we will use Fubini's formula. So I have to prove that
$$
\int_0^1\int_0^{1-x_n}…\int_0^{1-(x_n+…+x_2)}1\,\,\,dx_1…dx_{n-1}dx_n=\frac 1{n!}
$$
but unfortunately I don't be able to prove it.
For sake of completeness I point out that it seems that here there is a similar solution to that I gave (see the answer of the professor Blatter) but I don't fully understand it. In particular the solution I linked says that if we define
$$
E_\xi:=\{x\in\Bbb R^n:x_1+…+x_{n-1}\le1,\,\,\,\text{and}\,\,\,x_1,…,x_{n-1}\ge 0\,\,\,\text{and}\,\,\,x_n=\xi\}
$$
for any $\xi\in[0,1]$ then $E=\bigcup_{\xi\in[0,1]}E_\xi$ and so if we observe that the projection of $E_\xi$ is a $(n-1)$ dimensional simplex then $\int_E 1=\int_0^1(1-x_n)^{n-1}v(E_\xi)\,dx_n=\frac 1 n(1-x_n)^nv(E_\xi)=\frac 1{n!}$ that complete the proof but I don't understand how to prove effectively last equality.
So I ask to prove the last equality and then I ask to prove that $h[E]=S$ too. So could someone help me, please?
Best Answer
With a slight generalization, it becomes easy to prove by induction that for each $a$, we have: $$\int_0^a \int_0^{a-x_n} \cdots \int_0^{a-(x_n+\cdots+x_2)} dx_1 \cdots dx_n = \frac{a^n}{n!}.$$ The base case $n = 1$ is trivial. For the inductive case, we have using the inductive hypothesis that the desired integral is equal to $\int_0^a \frac{(a-x_n)^{n-1}}{(n-1)!} dx_n = \left. -\frac{(a-x_n)^n}{n!}\right|_{x_n=0}^{x_n=a} = \frac{a^n}{n!}$.
(And in fact, via the transformation that you outlined in the question, the general case given above can be proved from the special case $a=1$. This would give another way to proceed in a proof by induction: use the transformation along with the inductive hypothesis to conclude that the function being integrated with respect to $x_n$ is $\frac{(1-x_n)^{n-1}}{(n-1)!}$.)