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why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$

Any hints that can take me from here or am I completely lost.

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(\sum_{b=0}^{a-1}\binom{a}{b}(a-b)^n(-1)^b)(\sum_{a1

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(\sum_{b=0}^{a-1}\binom{a}{b}(a-b)^n(-1)^b)(\sum_{i=1}^{n}(n+1-i)^{a-1}(i)))$

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(\sum_{b=0}^{a}\binom{a}{b}(a-b)^n(-1)^b)(\sum_{i=1}^{n}(n+1-i)^{a-1}(i)))$

$\sum_{k=1}^{n}{k^p}=\sum_{a=1}^{p}(-1)^{p-a}(a!S(n,a))(\sum_{i=1}^{n}(i)^{a-1}(n+1-i)))$

Where S(n,a) is a stirling number of second kind.

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    I assume $p$ is a positive integer. There is a moderately complicated [General Formula](http://en.wikipedia.org/wiki/Faulhaber%27s_formula) for the sum, involving the Bernoulli numbers. The term "power series" is a technical term that refers to something else.2012-04-12
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    Yes p is a positive integer. Thanks for correcting me.2012-04-12
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    If you like you look at http://go.helms-net.de/math/potenzsummen/potenzsummen_1.htm which is a very old and completely elementary treatize of mine, of when I explored this myself the first time. Unfortunately in german, but I think the formulae and expressions are clear enough to hint you to a fruitful direction (for instance introducing how the Eulerian numbers of hkju's answer come into play)2012-04-12

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