Find the volume formula of a simplex proving that $\int_0^1\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_n=\frac 1{n!}$ 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 $i&gt;0$.

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 $n\in\Bbb N$.
So if $x_0,...,x_n$ are $(n+1)$ point affinely independet 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 unfornuately 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?
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 $\frac{(1-x_n)^{n-1}}{(n-1)!}$.)
First of all I summarise to follow what I above defined to point out some things.
Definition
If $x_0,...,x_n$ are $(n+1)$ affinely independent point of $\Bbb R^n$ then the simplex determinated by them is the set
$$
S:=\Biggl\{x\in\Bbb R^n: x=x_0+\alpha_1v_1+...+\alpha_nv_n,\,\sum_{i=1}^n\alpha_i\le1\,\,\,\text{and}\,\,\,\alpha_i\ge 0\,\,\,\text{for all}\,i\Biggl\}
$$
where $v_i:=(x_i-x_0)$ for each $i&gt;0$.
So with this definition it is clear that a simplex can be always individuated by one point $x_0$, that we call origin, and by vectors of a base $v_1,...,v_n$, that we call directions. Finally we call the points $(x_0+v_1),...,(x_0+v_n)$ vertex.
So let's start solve the question.
Theorem
If $S$ is a simplex generated by the affinely independent points $x_0,...,x_n$ of $\Bbb R^n$ then
$$
v(S)=\Big|\frac{1}{n!}\det\big[v_1,...,v_n\big]\Big|
$$
where $v_i:=(x_i-x_0)$ for each $i=1,...,n$.
Proof. So if $x_0,...,x_n$ are $(n+1)$ affinely independent points then we define the tranformation $h:\Bbb R^n\rightarrow\Bbb R^n$ through the condition
$$
h(x):=A\cdot x+x_0
$$
where $A$ is the matrix whose $j$-th column is for each $j=1,...,n$ the vector $v_j$ as above defined. First of all we observe that the transformation $h$ is a diffeomorphism because the vectors $v_1,...,v_n$ are linearly independet so that the matrix $A$ is not singualr. Then we observe that the trasformation $h$ map the simplex $E$ with origin the intersection $O$ of the axes and with directions the vector of canonical base $\mathcal E:=\{e_1,...,e_n\}$ onto the simplex $S$ because $h$ map the vertex of $E$ in the vertex of $S$. Now we suppose that
$$
v(E)=\frac 1{n!}
$$
so that we observe that in this particular case from the change of variables theorem it would follow that
$$
v(S)=\int_S 1=\int_E |\det A|=|\det A|\int_E 1=|\det A|v(E)=\frac 1{n!}|\det A|
$$
and so the theorem would hold. So let's prove the theorem using induction. So clearly if $n=1$ then $S=[a,b]$ for some $a,b\in\Bbb R$ and so the formula trivially holds. So we suppose that the formula holds for $(n-1)$ and we prove that it holds of $n$. So we know that it is sufficient to prove that
$$
\int_E 1=\int_0^1\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_{n-1}dx_n=\frac 1{n!}
$$
and exactly we can do this knowing that the theorem holds for any siplex of $\Bbb R^{n-1}$ by the inductive hypotesis. So fixing $x_n$ let be $\xi:=(1-x_n)$ and we define
$$
T:=\big\{x\in\Bbb R^{n-1}:x_1+...+x_{n-1}\le\xi\,\,\,\text{and}\,\,\,x_i\ge 0\,\,\,\text{for all}\,\,\,i=1,...,(n-1)\big\}
$$
that is $T$ is the simplex of $\Bbb R^{n-1}$ with origin the intersection $O$ of axes and with directions  those vectors that are proportional to the vectors of canonical base and of length $\xi$. So by the inductive hypotesis we know that
$$
v(T)=\Biggl|\frac 1{(n-1)!}\det[\xi e_1,...,\xi e_n]\Biggl|=\frac 1{(n-1)!}(1-x_n)^{n-1}
$$
and moreover by definition of volume
$$
v(T)=\int_T 1=\int_0^{1-x_n}...\int_0^{1-(x_2+...+x_{n-1}+x_n)}1dx_1...dx_{n-1}
$$
so that we conclude that
$$
\int_0^1\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_{n-1}dx_n=
\\
\int_0^1\Biggl[\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_{n-1}\Biggl]dx_n=
\\
\int_0^1\frac 1{(n-1)!}(1-x_n)^{n-1}dx_n=-\frac 1{n!}(1-x_n)^n\Biggl|_0^1=\frac 1{n!}
$$
and so the theorem holds.