For $p$ prime, is there a simple expansion for the Stirling number of the second kind $S(n,k)$, of the form $S\left(n,\frac{p-1}{2}\right) \pmod{p}$? I tried using the explicit expansion $$\left( \frac{p-1}{2} \right)!S(n,\frac{p-1}{2}) = (-1)^{\frac{p-1}{2}}\sum_{l=0}^{\frac{p-1}{2}}(-1)^l \binom{\frac{p-1}{2} }{l}l^n$$ and a literature search, but got nowhere.
For context: the von Staudt - Clausen theorem exactly describes the fractional part of the $n$-th Bernoulli number $B_n$, by writing $B_n$ as a sum over Stirling numbers and reducing mod $p$. This means that we have to analyze $S(n,p) \pmod{p}$, which is either $0$ or $1$. I'm looking at a specialization of poly-Bernoulli numbers, which are multiple sums over Bernoulli numbers that arise when we try to generalize the formula for the Riemann zeta function $\zeta(2n) = (-1)^{n+1} \frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$. Finding an analog of the von Staudt - Clausen theorem ends up reducing to analyzing the Stirling numbers $S\left(n,\frac{p-1}{2}\right) \pmod{p}$, and an explicit expression would give us some arithmetic information about a special kind of multiple zeta function.
Numerically, it looks like $S\left(n,\frac{p-1}{2}\right) \pmod{p}$ is periodic with period $p-1$, but I still can't conjecture the actual value.