Consider the set
$$A(\alpha):=\left\{ x \in \mathbb{R}^n: 0\le x_1\le x_2 \le...\le x_n \le \alpha \right\}$$
is it then possible to determine the volume of this set?
After some trying I obtained a bound on the volume of this set that reads
$$\left\lvert A(\alpha) \right\rvert \le \frac{\alpha^n}{n!},$$
but I did not manage to calculate the volume exactly. Is this possible at all?
The volume is given by $$\int_0^{\alpha}\int_{x_1}^{\alpha}\cdots\int_{x_{n-1}}^{\alpha}dx_n\cdots dx_2dx_1$$ Starting with the inner integral $$\int_{x_{n-1}}^{\alpha}dx_n=\alpha-x_{n-1},$$ then $$\int_{x_{n-2}}^{\alpha}\int_{x_{n-1}}^{\alpha}dx_n dx_{n-1}=\frac{(\alpha-x_{n-2})^2}{2},$$ and $$\int_{x_{n-3}}^{\alpha}\int_{x_{n-2}}^{\alpha}\int_{x_{n-1}}^{\alpha}dx_n dx_{n-1}dx_{n-2}=\frac{(\alpha-x_{n-3})^3}{6}.$$ By induction, it then follows that $$\int_0^{\alpha}\int_{x_1}^{\alpha}\cdots\int_{x_{n-1}}^{\alpha}dx_n\cdots dx_2dx_1=\frac{\alpha^n}{n!}.$$