Is it possible a closed expression for $1^k + 2^k + ... + n^k$? If so, am I on the way?

Question

I was asked to calculate the sequence $1^k + 2^k + ... + n^k$ and if it is possible to find a closed expression for it for every $k \in \mathbb{N}$. I could find that I can rewrite $\sum_{i=1}^ni^k$ into $1^k + \sum_{i=1}^{n-1} (1+i)^k$. Also it seemed to me that I could without prejudice use the superior summation limit as $n$, so that I would have a sum of binomial expressions (using $(n-1)$ the last term would be $n^k$ and for that I was unsure if I could use the binomial theorem within the sum). Rewriting $1^k + \sum_{i=1}^{n} (1+i)^k$ again (using the mentioned theorem) I got: $1 + \sum_{i=1}^{n} \sum_{j=0}^k {k \choose j}i^j$, but now I'm stuck. Any tips? Am I on a reasonable course?

Answer 1

Hint: You could find in this answer formulas for the sum of $k$-th powers of the first $n$ natural numbers.

\begin{align*} \sum_{j=1}^nj^k&=[z^n]\frac{1}{1-z}(zD_z)^k\frac{1}{1-z}\\ &=\sum_{j=1}^k{k\brace j}\frac{(n+1)^{\underline{j+1}}}{j+1} \end{align*}

with $D_z:=\frac{d}{dz}$ the differential operator and ${k\brace j}$ Stirling numbers of the second kind.