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.

You can use the closed-form expression from Wikipedia:

$$A(n,m)=\sum_{k=0}^{m+1} (-1)^k\dbinom{n+1}{k}(m+1-k)^n$$

At $k=0$ the binomial is $1$, and Fermat Little Theorem tells us the last expression is $1\pmod{p}$. All the other binomials are divisible by $n+1=p$.