A formula for stirling number of first kind

Question

The recurrence gives Stirling number of the first kind is: $$s(r,n) = s(r - 1, n - 1) + (r - 1)s(r - 1, n) \qquad (n \ge 1)$$ However, I noticed that $s(5,2) = 50 = (1\cdot2\cdot3) + (1\cdot2\cdot4) + (1\cdot3\cdot4) + (2\cdot3\cdot4)$, numbers in the parenthesis are all the $(5-2) = 3$ element subsets of the set of numbers from $1$ to $(5-1) = 4$ . This seems to hold for any $s(r,n)$ . Is there any way to prove this? How can I translate this idea into mathematical notation? I worked via induction, assuming it holds for some $k \ge 2$ and then proving for $s(k+1,n) = s(k, n - 1) + k\cdot s(k , n)$ . I am quite a novice and can't figure out how to write the induction proof and whether this is correct.

Answer 1

OP's claim is that

$${n\brack n-k} = [z^k] \prod_{q=1}^{n-1} (1+qz).$$

Now the OGF of unsigned Stirling cycle numbers is

$$n! \times {z+n-1\choose n}.$$

For a polynomial $p(z)$ of degree $n$ the reversed polynomial is given by $z^n p(1/z)$ so we get

$$[z^k] z^n \times n! \times {1/z+n-1\choose n} = [z^k] \prod_{q=0}^{n-1} (1+qz) = [z^k] \prod_{q=1}^{n-1} (1+qz)$$

which is the claim.