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.

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

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