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For the Stirling numbers of the second kind, the following identity is well-known: \begin{align} S(n,l) = \sum_{k_1 + \cdots + k_l = n-l} 1^{k_1} 2^{k_2} \cdots l^{k_{l}}, \end{align} where the sum is taken over non-negative integers satisfying $0 \leqslant k_1 \leqslant \cdots \leqslant k_l \leqslant n- l$. Is an analogous identity known for $s(n,l)$, the Stirling numbers of the first kind?

Edit: Thanks to Raymond, the correct formula is \begin{align} s(n,l) = (-1)^{n-l} \mathop{\sum_{k_1 + \cdots +k_{n-1} = n-l}}_{0 \leqslant k_i \leqslant 1} 1^{k_1} 2^{k_2} \cdots (n-1)^{k_{n-1}}. \end{align}

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Yes (26.8.3 of DLMF) :

$$(-1)^{n-k}s(n,k)=\sum_{1\leq b_1<\cdots<b_{n-k}\leq n-1} b_1 b_2\cdots b_{n-k} $$

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