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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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    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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$\sum_{k=1}^n k^p = \sum_{k=0}^{p-1} T(p,k)\binom{n+1+k}{p+1}. \mbox{ (hence, its degree is $p+1$)}$, where $T(p,k)$ is the Eulerian number (cf.OEIS A008292). For example, $p=2$:$\sum_{k=1}^n k^2 = \sum_{k=0}^{1} T(2,k)\binom{n+1+k}{3}=(1)\binom{n+1}{3}+(1)\binom{n+2}{3}=\frac{n(n+1)(2n+1)}{6}$ $p=3$:$\sum_{k=1}^n k^3 = \sum_{k=0}^{2} T(3,k)\binom{n+1+k}{4}=(1)\binom{n+1}{4}+(4)\binom{n+2}{4}+(1)\binom{n+3}{4}=\frac{n^2 (n+1)^2}{4}$