# Nested sum $\sum_{i<j< \cdots < k} ij \cdots k$

I am wondering if there is any known closed form for the following nested sum? :

$$\sum_{i<j<\cdots <k} ij\cdots k$$ where each $i,j,\cdots,k =1, \cdots, n$

I tried the first one:

$$\sum_{i<j}ij = \sum_{i=1}^n i \sum_{j=1}^{i-1}j = \sum_{i=1}^n i \binom{i}{2} = \frac{n(n+1)}{2}\left[\frac{n(n+1)}{2} - \frac{1}{2} \right]$$ the next one looks worse, it is the sum of the above making $n \to k-1$ and summing over $k=1$ to $n$.

These are the absolute values of the Stirling numbers of the first kind. If you write down the polynomial $$(x-1)(x-2)\cdots (x-n)$$ then the coefficient of $x^{n-r}$ is (up to sign) the sum of the products of distinct $r$-tuples of the numbers from $1$ to $n$, which are the numbers you are after. See the OEIS entry for lots of information.