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Let $S_{n,r} := \sum_{k=1}^{n} k^r$ be the power sum. On the homepage by W. Hecht (link) I have found the following determinant expression:

$$S_{n,r} = (-1)^{r-1} \frac{n(n+1)}{(r+1)!} \det \begin{pmatrix} 2 & 0 & 0 & & \cdots && 1 \\ \binom{2}{0} & 3 & 0 & & \cdots && -n \\ \binom{3}{0} & \binom{3}{1} & 4 & 0 & \cdots && n^2 \\ \vdots & & & & && \vdots \\ \binom{r}{0} & \binom{r}{1} & \binom{r}{2} & & \cdots & \binom{r}{r-2} & (-1)^{r-1} n^{r-1} \end{pmatrix}$$

I quite like this explicit formula, but I wonder (and actually W. Hecht too) if this has ever appeared in the literature so far.

Here is a sketch of the proof which you can find on the homepage: For every $r$ one verifies the (well-known, right?) recurrence formula

$$\sum_{j=1}^{r} (-1)^{j-1} \binom{r}{j-1} S_{n,j} + (-1)^{r-1} S_{n,r} = (-1)^{r-1} n^r (n+1)$$

Interpret this as a system of linear equations and apply Cramer's rule.

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+1 Quite an interesting identity. Although if you're looking for a formula to compute power sums, the RHS is more complicated than the sum it's computing. :) – Bill Cook Mar 14 '12 at 1:26
@Bill: The RHS is an explicit polynomial of degree $r+1$ in $n$. This is how one usually computes the LHS (plug in $r=1,2,3,\cdots$ and see what happens). – Martin Brandenburg Mar 14 '12 at 16:06

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