I have seen a proof that the recurrence of the unsigned stirling numbers of the first kind can be derived from the rising factorial definition.
That is, $s(n,k)$ are given by the coefficients on $x^k$ in the expansion $x(x+1)...(x+n-1)$ and this definition can be used to show that $s(n,k)=(n-1)s(n-1,k)+s(n-1,k-1)$
This seems to be the general progression. Start with the rising factorial definition, and then it is fairly straightforward to show the recurrence relation.
My Question:
Is there a good way to see this in reverse? Can you start with the stirling recurrence I have listed above and show that the stirling numbers must come from the coefficients of the rising factorial polynomial?
Sure, just call $$f_n(x)=\sum _{k=0}^ns(n,k)x^k,$$ use the base case to understand $f_0(x)$ and then, for $n>0$, just plug the recurrence as $$f_n(x)=\sum _{k=0}^ns(n,k)x^k=\sum _{k=1}^{n}\left ((n-1)s(n-1,k)+s(n-1,k-1)\right )x^k,$$ split the sum and conclude that $f_n(x)=(n-1)f_{n-1}(x)+xf_{n-1}(x)=f_{n-1}(x)(x+n-1).$
Show that the rising factorials have the same recursion and you are done.