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The closed form formula for $\sum_{r=1}^n(r^k)$ can be expressed as a polynomial of degree $k+1$, for example $\sum_{r=1}^n(r^2) = \frac{1}{6}n(n+1)(2n+1)$.

This seems to hold true for all values of $k$. But why can this sum be expressed as a polynomial? Is there a proof (which doesn't rely on Faulhaber's formula) that this sum can always be expressed as a degree $k+1$ polynomial?

S34NM68
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    Why would you exclude Faulhaber? Anyway, it's easy to argue that the finite difference map takes polynomials of degree $≤d$ to those of degree $≤d-1$ and that the kernel is precisely the constants. Hence it must be surjective. – lulu Apr 01 '22 at 23:31
  • See this answer for an elementary demonstration that uses only Algebra and induction. It avoids Calculus, Bernoulli Numbers and Faulhaber. – user2661923 Apr 02 '22 at 00:09
  • https://math.stackexchange.com/questions/1893693/elementary-proof-of-polynomial-degree-of-sum-of-pth-powers?rq=1 I just found this post which answers the same question. – S34NM68 Apr 02 '22 at 15:29

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