Dividing standard simplex in equal parts

Question

Take the standard simplex $\Delta^n=\{x_1+\ldots+x_{n+1}=1,x_i\geq 0\;\forall i=1,\ldots,n+1\}$. Consider $\Delta^1$, I can divide it into two equal parts by considering the two sets $\{x\in\Delta^1:x_1<x_2\}$ and $\{x\in\Delta^1:x_1> x_2\}$. Consider $\Delta^2$, I can divide it into six equal parts by considering $\{x\in\Delta^2:x_3<x_2<x_1\}$, $\{x\in\Delta^2:x_3<x_1<x_2\}$ etc going for all the permutations on three letters (all $\sigma\in S_3$ that is). Is this true in general for any dimension? That is, is it true that I can divide $\Delta^n$ into $(n+1)!$ equal parts by using the sets $\{x\in\Delta^n:x_{\sigma(1)}<\ldots<x_{\sigma(n+1)}\}_{\sigma\in S_{n+1}}$ ? How would one prove it?