Let $A(m,n)$ denote the [Eulerian numbers](https://en.wikipedia.org/wiki/Eulerian_number).
I'm looking for a simple combinatorial proof of the following fact.



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> **Fact.** If $p$ is prime and $0\le k < p-1$, then $A(p-1,k) \equiv 1 \pmod{p}$.
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The closest thing I'm aware of is an argument of S. Tanimoto, An operation on permutations and its application to Eulerian numbers, *European Journal of Combinatorics* **22** (2001), 569–576 that can be adapted to give a rather complicated proof, but I'm hoping for something simpler and more direct.



Combinatorial proof of fact about Eulerian numbers?
We identify the permutations of $1,2,\dots,p-1$ and the cyclic permutations $c=(c\_0,\dots,c\_{p-1})$ of $0,\dots,p-1$: if $c\_k=0$, $c$ corresponds to $\pi(c):=(c\_{k+1},c\_{k+2},\dots,c\_{k-1})$. There are $p-1$ cyclic permutations which are arithmetic progressions $(0,a,2a,\dots,(p-1)a)$, and other cyclic permutations are partitioned onto orbits of size $p$ of the action of $\mathbb{Z}\_p$: $(c\_0+a,c\_1+a,\dots,c\_{p-1}+a)$. The key fact is that in each group $C$ the corresponding permutations $\pi(c)$, $c\in C$, have the same number of ascents (example: for a cyclic permutation $01324$ the orbit contains also $12430$, $23041$, $34102$, $40213$, and permutations of $1234$ which are $1324$, $1243$, $4123$, $2341$, $2134$, all have 2 ascents.) The result follows after we also establish that the arithmetic progression $(a,2a,\dots,(p-1)a)$ (multiplication is taken modulo $p$) has exactly $p-1-a$ ascents. 


