User:Mirceamrk/sandbox/Faulhaber's formula (→Relationship to Stirling numbers)

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Relationship to Stirling numbers
Using the unsigned Stirling numbers of the first kind $$ \left[{n\atop k}\right] $$ and the Stirling numbers of the second kind $$ \left\{ {n\atop k} \right\} $$, one can write :

\sum\limits_{k=1}^n k^p = \sum\limits_{k=1}^p (-1)^{k-1} k \left[{n+1\atop n+1-k}\right] \left\{ {n+p-k\atop n} \right\}. $$