Past days I've been trying to prove that certain polynomials have positive coefficients. After a lot of thinking, I came up with a formula for each coefficient individually, and they are not *that* ugly. I think maybe someone can give me a hand on this. 

Synthetically, what I want to prove is that the following sum is positive:

$$S(k,n,m)=\sum_{i=0}^{n-m-1} \sum_{j=0}^{k-1} (-1)^{i+j} \binom{n}{j}(k-j)^m {j \brack {j-i}} {{n-j}\brack {m+1+i-j}}$$

Where the symbol ${x \brack y}$ stands for the Stirling numbers of the first kind (without sign).

I'm interested in the case $1\leq m,k\leq n-1$.

I have already proven the following:

1) If in the sum we set $m=n-1$, we get just the well known recurrence for Eulerian numbers, so it is positive. For $m=n-2$, the result is a sum of two Eulerian numbers.

2) If we replace $k$ by $n-k$, the sum remains the same (the proof of this fact is somewhat abstract in the sense one has to understand the context on which this sum arises).

3) With $k=1$, we get simply the Stirling numbers of the first kind.

4) With $m=1$ the sum is always positive. 

I don't mind if the proof is strictly combinatoric or involves inequalities of Stirling numbers or even uses the exponential generating function of some of the numbers inside. However, I strongly suspect that this alternating sum can be rewritten as a sum of products of stirling, eulerian, or binomial numbers (I couldn't manage to guess such a formula in any case not listed before).

I posed this question in MSE, but I think it fits a lot better here