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 : 

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

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