I'm trying to find an explicit number theory definition of the Stirling numbers of the first kind in reverse as $|s(n,k)|$ not $|s(n,n-k)|$, not the way it's usually written. Here is the typical definition in most books
$$ |s(n,n-k)|=\prod_{l=1}^{k} \sum_{i_l=0}^{i_{l-1}-1} i_l , \quad i_o=n $$
Or alternatively
$$ |s(n,n-k)|=\sum_{0 \leq i_1 < i_2 < ... < i_k < n} i_1 i_2 ... i_k =\sum_{i_1=0}^{n-1}i_1\sum_{i_2=0}^{i_1-1}i_2...\sum_{i_k=0}^{i_{k-1}-1}i_k $$
In Gould's book on Stirling numbers, he cites a reverse formula, but it doesn't really enumerate into the same result, or seems either shifted or incorrect $$ |s(n,k)|=\prod_{l=1}^{k} \sum_{i_l=i_{l+1}+1}^{n-l+1} i_l , \quad i_{k+1}=0 $$
Or
$$ |s(n,k)|=\sum_{i_k=1}^{n-k+1} i_k \sum_{i_{k-1}=i_k+1}^{n-k+2}i_{k-1}...\sum_{i_2=i_3+1}^{n-1} i_2 \sum_{i_1=i_2+1}^{n}i_1 $$
where it is known that for some index $l$ and positive integers $s,m$
$$ \sum_{k=l+s}^{m} k=-(\sum_{k=0}^{l+s-1} k - \sum_{k=0}^{m} k) $$
For example, this result for $|s(n,k)|$ in the case of $n=5,k=2$ gives
$$ \sum_{i_2=1}^{4} i_2 \sum_{i_1=i_2+1}^{5} i_1 = \sum_{i_2=1}^{4} i_2(-{i_2 \choose 2}+15)=85=|s(6,4)| $$
I'm not trying to find an easy way to compute Stirling numbers, or use recurrence relations on a computer. The question is about specifically the formula for $|s(n,k)|$ any help would be appreciated