I'm confused about the identity of Stirling numbers: $s(n,k) = k!S(n,k)$

Question

I'm a little bit confused about this identity between Stirling numbers of the first and second kind, because from it you can derive that:

$s(n,k) = k!S(n,k) = k!(S(n-1,k-1) + kS(n-1,k)) = k(k-1)!S(n-1,k-1) + kk!S(n-1, k) = ks(n-1,k-1) + ks(n-1,k) = k(s(n-1,k-1) + s(n-1, k))$.

But the identity for Stirling numbers of the first kind is known and it's

$s(n,k) = s(n-1,k-1) + (n-1)s(n-1,k)$.

Have I mistaken somewhere or both identites are correct?

Answer 1

1. If you define Stirling numbers of the first kind as

   The number of permutations of $\{1,2,\dots,n\}$ with exactly $k$ cycles,

   then the equation $s(n,k)=s(n-1,k-1)+(n-1)s(n-1,k)$ is correct.

2. If you define Stirling numbers of the first kind as

   The number of surjective functions from a set of size $n$ to one of size $k$,

   them the equation $s(n,k)=k\big(s(n-1,k-1)+s(n-1,k)\big)$ is correct.

These two cannot be correct simultaneously.

Your confusion results from the fact you are using the same symbol, $s(n,k)$, for two different quantities.