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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 $n$.

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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.

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A few first formulas as polynomials of even order:
\begin{array}{lrl} S_1(n) = \displaystyle\sum_{1\le i \le n} i = & \frac{1}{2}\;n & (n+1), \\ S_2(n) = \displaystyle\sum_{1\le i<j \le n} ij = & \frac{1}{24}(n-1)\;n & (n+1)(3n+2), \\ S_3(n) = \displaystyle\sum_{1\le i<j<k \le n} ijk = & \frac{1}{48}(n-2)(n-1)\;n & (n+1)(n^2+n), \\ S_4(n) = & \frac{1}{5760} (n-3)(n-2)(n-1)\;n & (n+1)(15n^3+15n^2-10n-8), \\ \cdots \end{array}

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