Computing $n!\int...\int_{0Compute $n!\int...\int_{0<s_1<...<s_n}ds_1...ds_n$. 
Before that, I were to show that $$\int...\int_{0<s_1<...<s_n}f(s_1,s_2-s_1,...,s_n-s_{n-1})ds_1...ds_n=\underset{0<x_1,...,x_n \\ x_1+...+x_n<1}{\int...\int}f(x_1,x_2,...,x_n)ds_1...ds_n $$
which I did using a change of variable theorem. 
I don't know, however, how to apply it here, if I better apply here. I generally don't know what the bounds for each integral are. I could really use some help.
 A: Assuming that $X_1,X_2,\ldots,X_n$ are independent random variables, uniformly distributed over $(0,1)$, the integral
$$ \int_{0\leq x_1\leq x_2 \leq \ldots\leq x_n\leq 1} 1\,d\mu $$
computes the probability that $X_1\leq X_2\leq\ldots\leq X_n$, that is obviously $\frac{1}{n!}$, since
$$(x_1,x_2,\ldots,x_n)\quad\text{and}\quad (x_{\sigma(1)},x_{\sigma(2)},\ldots,x_{\sigma(n)}) $$
have the same probability to occur for any $\sigma\in S_n$, and $|S_n|=n!$.
That proves that the value of your integral is just $\color{red}{\large 1}$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Any permutation of the dummy variables $\ds{\braces{s_{1},\ldots,s_{n}}}$ leaves the integral unchanged. Because there are $n!$ permutations of them, we can write ( the sum $\sum_{P}$ is over the above mentioned possible permutations and $\Theta$ is the Heaviside Step Function )
\begin{align}
&\color{#f00}{%
n!\int\cdots\int_{0\ <\ s_{1}\ <\ \cdots\ <\ s_{n}\ < 1}
\dd s_{1}\cdots\dd s_{n}}
\\[3mm] = &\
n!\bracks{{1 \over n!}\sum_{P}\int_{0}^{1}\cdots\int_{0}^{1}
\Theta\pars{s_{1}}\Theta\pars{s_{2} - s_{1}}\cdot\Theta\pars{s_{n} - s_{n - 1}}
\dd s_{1}\cdots\dd s_{n}}
\\[3mm] = &\
\int_{0}^{1}\cdots\int_{0}^{1}\overbrace{\bracks{%
\sum_{P}\Theta\pars{s_{1}}\Theta\pars{s_{2} - s_{1}}\cdot
\Theta\pars{s_{n} - s_{n - 1}}}}^{\ds{=\ 1}}\ \,\dd s_{1}\cdots\dd s_{n} =
\color{#f00}{1}
\end{align}
In $\underline{\mbox{Many-Body Physics}}$, it's usually written as
$$
\mathcal{T}\int_{0}^{1}\cdots\int_{0}^{1}\dd s_{1}\,\dd s_{n} =
\mathcal{T}\pars{\int_{0}^{1}\dd s}^{n}
$$
where $\mathcal{T}$ is the Dyson Chronological Operator.
