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Let $s(n,k)$ denote the signless Stirling numbers of the first kind. Prove that:

$$s(n,2) = (n-1)!(1 + \frac{1}{2} + \frac{1}{3} +...+ \frac{1}{n-1})$$

-I haven't dealt with Taylor series expansion in a long time and not quite sure how (brand new to)Stirling numbers would play into this proof.Any help is appreciated.

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HINT: If you split $[n]$ into two cycles, one of length $k$ and the other of length $n-k$, there are $\binom{n}k$ ways to choose the elements of the $k$-cycle, $(k-1)!$ ways to arrange them in a cycle, and $(n-k-1)!$ ways to arrange the remaining elements in a cycle. That gives you a total of

$$\binom{n}k(k-1)!(n-k-1)!$$

permutations. Now sum over the possible values of $k$. Be careful, though: you’ll be counting every permutation twice.

You will probably also find it useful to notice that

$$\frac{n}{k(n-k)}=\frac1k+\frac1{n-k}\;.$$

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