let $ 1\leq n $ and $a_1,...,a_n \mathbb{R}^{+} $
A simplex $ \sigma $ ist given by $ \sigma := [0,a_1 e^1, .., a_n e^n ] $
I want to proove that
$$ \int_ {\sigma } 1= \frac{1}{n!} \prod_{j=1}^n a_j $$
per induction.
first, I wanted to clarify the formula for lower n.
for $ n=1 , [0,e^1]$ you get $\int_{[0,e^1] }1= \int_0^{e^1} dx= e^1$ and set $e^1 = 1$
for $n=2, [0,e^1,e^2]$ you get $\int_{[0,e^1,e^2]} 1 = \int_0^{e^1} \int_0^y dx dy = \frac{(e^1)^2}{2} $
for $ n=3, [0,e^1,e^2,e^3] $ you get, that $ \int_{[0,e^1,e^2,e^3]} 1= \int_0^{e^1} \int_0^z \int_0^y dxdydz= \frac{(e^1)^3}{6} $
my question is also, if this is formally right? fells kinda wrong
so
$\int_{[0,e^1,e^n,e^{n+1}]} 1 = \int_0^{e1} \int_0^{x_{n+1}}....\int_0^{x_2} dx_1...dx_n dx_{n+1} $
$= \int_0^{e^1} \frac{1}{n!} x_{n+1}^n dx_{n+1} = \frac{e^{n+1}}{(n+1)!} $ and also set $ e^{n+1}=1$ ..again..is this formally correct?
My two cents : first do the linear, diagonal change of variables $y_i=a_i * x_i$. The determinant of the jacobian matrix of this transformation is uniform, and its value is $\prod_i a_i$. The pre-image of your simplex is then the so-called "standard" simplex $\sigma_0 = [0,e^1,\dots,e^n]$, so that all you are left to prove is $\int_{\sigma_0} 1 = \frac{1}{n!}$. This is done using recurrence on the dimension $n$.
Edit : This is exactly what was already suggested in the comments, only a tiny bit more verbose.