a polytope, so in particular $E^*(\mathcal{S}_\sigma,z)$ is well defined, and we have:

\[ehrssigma\] $E^*(\mathcal{S}_\sigma,z) = z^{\operatorname{des}(\sigma^{-1})}$.

We can check that $v \in [0,1]^n$ is in $\mathcal{S}_{\sigma}$ if and only if $v_{\sigma^{-1}(1)} \leq \dots \leq v_{\sigma^{-1}(n) }  $, and $v_{\sigma^{-1}(i)} < v_{\sigma^{-1}(i+1) }$ if $\sigma^{-1}(i)>\sigma^{-1}(i+1)$. The number $E(\mathcal{S}_{\sigma},t)$ counts such sequences with the additional condition that all elements are integers between $0$ and $t$. By defining $$w_i = v_{\sigma^{-1}(i)} - \operatorname{des}( \sigma^{-1}(1),\dots, \sigma^{-1}(i) ),$$ we have a bijection with integer sequences satisfying $0\leq w_1 \leq \dots \leq w_n \leq t-\operatorname{des}(\sigma^{-1})$, so that $$E(\mathcal{S}_{\sigma},t) = \binom { n + t - \operatorname{des}(\sigma^{-1}) }{n}.$$ The expansion $$\sum_{t\geq0 } \binom tn z^t = \frac{z^n}{(1-z)^{n+1} }$$ permits to finish the proof.

The [*Eulerian polynomials*]{} are $$A_n(z) := \sum_{\sigma \in \mathfrak{S}_n } z^{ \operatorname{des}(\sigma^{-1} ) } = \sum_{\sigma \in \mathfrak{S}_n } z^{ \operatorname{des}(\sigma ) }.$$ Note that their coefficients are the numbers $A_{n,k}$ defined in Equation .

For example, $A_0(z)=A_1(z)=1$, $A_2(z)=1+z$, $A_3(z)=1+4z+z^2$, etc. Since the Ehrhart series is additive with respect to disjoint union, we get from the previous lemma that $$E^*( [0,1]^n , z ) =  A_n(z).$$ Note that, since the Ehrhart polynomial of $[0,1]^n$ is clearly $(t+1)^n$, we have proved the classical identity: $$\label{ideul}
  