$d$-dimensional set with Lebesgue measure $\frac{1}{d!}$ Let $X = \{(x_1, ..., x_d)\,|\,0 \leqslant x_1\leqslant ... \leqslant x_d\leqslant 1\}$. Show that $X$ has Lebesgue measure $\lambda_d(X)=\frac{1}{d!}$. I assumed that because of Fubini's theorem $$\lambda_d(X) = \int_X 1 dA = \int_0^1\int_{0}^{x_d}\int_{0}^{x_{d-1}}...\int_0^{x_2} 1 dx_1dx_2...dx_d$$ wich would work out to $\frac{1}{d!}$ due to the power rule. I noticed that the actual integral should be $$\int_0^1\int_{x_{d-1}}^{x_d}\int_{x_{d-2}}^{x_{d-1}}...\int_0^{x_2} 1 dx_1dx_2...dx_d$$
wich doesn't telescope or result in the actual measure. How do I show that $\lambda_d(X) =\frac{1}{d!}$?

All the $d!$ permutations  of the coordinates  give the same value and their sum is $\int_0^{1}\int_0^{1}...\int_0^{1} dx_1d_2...dx_d=1$.
