Thomas Wieder, Comments on A008275:

Let S=choose(n-1,n-k) be the set of all subsets with n-k elements (= combinations) 
taken from the set {1,2,...n-1}. 
E.g. for n=5 and k=2 we have S={{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}}. 
For each subset we form the product of its elements and sum over all these 
products and get the Stirling numbers of the first kind S1(n,k).

E.g. for n=5 and k=2 we have
(-1)^(n-k)*(1*2*3 + 1*2*4 + 1*3*4 + 2*3*4 = 6 + 8 + 12 + 24 = -50 = S1(5,2).

Let sum_{i=1, S=choose(n-1,n-k)}^{C(n-1,n-k)} denote a sum over all these 
combinations where the index i runs from 1 to the number of subsets. The later 
is equal to the binomial coefficient C(n-1,n-k). Furthermore, let s(i,j) be the 
j-th element of the i-th subset. Then

S1(n,k) = 
(-1)^(n-k)*sum_{i=1, S=choose(n-1,n-k)}^{C(n-1,n-k)} prod_{j=1}^(n-k) s(i,j).

A008275 Stirling1Number := proc(n::integer,k::integer)
local n1,k1,i,j,setlist,aset,S1,term;
# Procedure Stirling1Number calculates the Stirling numbers S1(n,k).
# http://www.thomas-wieder.privat.t-online.de/default.html
# Letzte Aenderung: 19.11.2006
n1:=n-1;
k1:=n-k;
setlist:=choose(n1,k1);
S1:=0;
for i from 1 to nops(setlist) do
aset:=op(i,setlist);
term:=1;
for j from 1 to nops(aset) do
term:=term*op(j,aset);
od;
S1:=S1+term;
od;
S1:=S1*(-1)^k1;
print(n,k,S1);
end proc;