<s>[INST] Formula for $1^k+2^k+3^k...n^k$ for $n,k \in \mathbb{N}$ So I've been looking for a formula where I can input the parameter $k$ and it will give me a formula for $1^k+2^k+3^k...+ n^k$ with $n,k \in \mathbb{N}$. The result is always a polynomial with $k+1$ as highest power. I've taken the time to calculate the polynomes for $k=1$ to $k=10$ by hand and using the interpolation feature of Wolfram Alpha. Here are the results (I'll only show the coefficients for the sake of clarity. the coefficients are always from $n^{k+1}$ to $n^1$. the constant is always $0$. So $\frac{1}{2},-\frac{1}{2}$ becomes $\frac{1}{2}n^2-\frac{1}{2}n$):


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*$k=1$ : $\frac{1}{2},-\frac{1}{2}$

*$k=2$ : $\frac{1}{3},\frac{1}{2},\frac{1}{6}$

*$k=3$ : $\frac{1}{4},\frac{1}{2},\frac{1}{4},0$

*$k=4$ : $\frac{1}{5},\frac{1}{2},\frac{1}{3},0,-\frac{1}{30}$

*$k=5$ : $\frac{1}{6},\frac{1}{2},\frac{5}{12},0,-\frac{1}{12},0$

*$k=6$ : $\frac{1}{7},\frac{1}{2},\frac{1}{2},0,-\frac{1}{6},0,\frac{1}{42}$

*$k=7$ : $\frac{1}{8},\frac{1}{2},\frac{7}{12},0,-\frac{7}{24},0,\frac{1}{12},0$

*$k=8$ : $\frac{1}{9},\frac{1}{2},\frac{2}{3},0,-\frac{7}{15},0,\frac{2}{9},0,-\frac{1}{30}$

*$k=9$ : $\frac{1}{10},\frac{1}{2},\frac{3}{4},0,-\frac{7}{10},0,\frac{1}{2},0,-\frac{3}{20},0$

*$k=10$ : $\frac{1}{11},\frac{1}{2},\frac{5}{9},0,1,0,1,0,-\frac{1}{2},0,\frac{5}{66}$


There are a few things i notice: Firstly, the coefficient of the highest power seems to be $\frac{1}{k+1}$. Secondly, the coefficient of the second highest power seems to be $\frac{1}{2}$ with the exception of $k=1$. Thirdly, all coefficients of the fourth, sixth, eight highest power and so on seem to be $0$. 
What is the formula that will output the coefficients for any value of $k$?
 [/INST] https://en.m.wikipedia.org/wiki/Faulhaber%27s_formula seems to deliver exactly what i needed. Original answer by MooS
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