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

Let $s(n,k)$ denote the unsigned Stirling numbers of the first kind. Prove that $$s(n, n-1) = {n\choose2}$$

-I am new to Stirling numbers, and looked up some definitions regarding it. I know that they count the number of permutations of n elements with k disjoint cycles. Although I am unsure of how to prove this and what kind of proof method to use.

• What does a permutation of $n$ things look like if it has $n-1$ disjoint cycles? – Matthew Towers Nov 7 '15 at 19:32
• I'm sorry i'm really not sure. – jeremysanchez50 Nov 7 '15 at 19:50
• Why not try n=3 first. Which permutations of $\{1,2,3\}$ have 2 disjoint cycles? For example, (1,2)(3) has 2 disjoint cycles but (1,2,3) only one. Then try n=4... – Matthew Towers Nov 7 '15 at 21:36
• so for $\{1,2,3\}$ there is (1,2)(3) (1,3)(2) and (2,3)(1)? – jeremysanchez50 Nov 7 '15 at 22:47
• Yes! What about n=4 and three disjoint cycles? – Matthew Towers Nov 7 '15 at 23:20