Is it true in general that $\int \dots \int_{0 \le x_1 \le \dots \le x_n,\ 0 \le x_n\le1}dx_1\dots dx_n=\left(\frac{1}{2}\right)^n?$ I was looking at an example with the following integral:
$$\iiiint_{0 \le x \le y \le z \le t,\ 0 \le t \le \frac{1}{2}} 1 \,dx\,dy\,dz\,dt = \frac{1}{16}$$
 Is it true in general that 
$$\int \dots \int_{0 \le x_1 \le \dots \le x_n,\ 0 \le x_n\le1}dx_1\dots dx_n=\left(\frac{1}{2}\right)^n?$$ 
(edit: here $0 \le x_n \le 1$ but in the example $0 \le t \le \frac{1}{2}$ so this can't be true...). Intuitively it would seem so since each dimension is "cut in half" by each inequality. Are there some other results for multiple integrals with this domain that have other integrand that is not a constant? Is it true that $$\int \dots \int_{0 \le x_1 \le \dots \le x_n,\ a \le x_n\le b}f(\boldsymbol{x})\,dx_1\dots dx_n=\int_a^b \int_0^{x_{n-1}} \int_0^{x_{n-2}} \dots \int_0^{x_2}f(\boldsymbol{x})\,dx_1\dots dx_n$$

To find the value of:
$$\int \dots \int_{0 \le x_1 \le \dots \le x_n,0 \le x_n\le1}dx_1\dots dx_n=?$$
First note that the first integral gives $x_1 |^{x_2}_0 = x_2$. The next one gives $x_3^2/2$ and so on.
This leads to: 
$$\int \dots \int_{0 \le x_1 \le \dots \le x_n,0 \le x_n\le1}dx_1\dots dx_n=\frac{L^{n-1}}{(n-1)!}$$
Where $L$ is the upper limit, in your case $1$, and in the original case, $1/2$.
