Show, for $p ∈ prime$ and $ 2 \le k \le p-1$
$\{\genfrac{}{}{0pt}{}{p}{k}\}$ is $divisible$ by $p$.
I'm looking for simple proof. I get that this problem was answered here:
Congruence for Stirling Number of first kind $s(n,k)$ when $n$ is prime
But if it's possible I'm looking for something simpler.
I saw Knuth approach to similar ($ 1 < k < p $) problem inside "Concrete Mathematics".
Knuth 6.51 a)
Is that proof lacking something?
As you mentioned, the idea is to consider the group action generated by adding $1$ to all of the elements of all the partitions, modulo $p$. The orbits of this action all have size $1$ or $p$, so you just need to conclude that size $1$ is impossible. Namely, for any set, let $f(A)$ denote the result of incrementing each element of $A$ by $1$ modulo $p$. You need to show that if $A_1,A_2,\dots,A_k$ and $f(A_1),\dots,f(A_k)$ are the same partition, then $k=1$ or $p$.
The idea is is show that these two partitions being the same implies all parts are the same size (which since $p$ is prime, implies these sizes must be $1$ or $p$). Indeed, for any $1\le i<j\le k$, to prove $|A_i|=|A_j|$, let $x$ be any element of $A_i$ and $y$ be any element of $A_j$. Letting $n=y-x\pmod p$, we see that $f^n(A_i)$ is a set which contains $y$. Since $f^n(A_i)$ is some set in the partition, it must be true that $f^n(A_i)=A_j$, as $A_j$ is the only set in the partition containing $y$. But $f$ preserves size of its input, so $|A_i|=|A_j|$.
The same logic applies to Stirling numbers of the first kind.