I discovered empirically that the pth (p is prime and greater than 3) Stirling number of the first kind is always divisible by $p^2$. If you calculate Stirling numbers recursively than the $p$th is given by: $$\left[(((1*2+1!)*3+2!)*4+3!)... \right]* (p-1)+(p-2)!$$
I think we should prove it similarly to the algebraic proof of Wilson's theorem but I don't know how exactly. Can you help me?
Example: $((1∗2+1!)∗3+2!)∗4+3!=50=2*25$
The Stirling number of the first kind that you are considering here is $\left[p\atop 2\right] =(p-1)!H_{p-1}$ where $H_{p-1}=1+\frac{1}{2}+..+\frac{1}{p-1}$ is a harmonic number.
This is not an algebraic proof, but it is shown here that $p^2$ divides $\left[p\atop 2\right] =(p-1)!H_{p-1}$ when $p\gt 3$ is prime.
This result is also known as Wolstenholme's theorem.