alternate form of sum of $r$th powers of the first $n$ positive integers

We will show that

$$\sum _{k=0}^n{k}^r={\int }_1^{n+1}{b}_r(x)𝑑x$$

We need two basic facts. First, a property of the Bernoulli polynomials is that $b_r^{\prime }(x)=r{b}_{r-1}(x)$. Second, the Bernoulli polynomials can be written as

$$b_r(x)=\sum _{k=1}^r\left(\genfrac{}{}{0pt}{}{r}{k}\right)B_{r-k}{x}^k+B_r$$

We then have

${\displaystyle {\int }_1^{n+1}}{b}_r(x)$$={\displaystyle \frac{1}{r+1}}(b_{r+1}(n+1)-b_{r+1}(1))={\displaystyle \frac{1}{r+1}}{\displaystyle \sum _{k=0}^{r+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{k}}\right)B_{r+1-k}({(n+1)}^k-1)$

$={\displaystyle \frac{1}{r+1}}{\displaystyle \sum _{k=1}^{r+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{r+1}{k}}\right)B_{r+1-k}{(n+1)}^k$

Now reverse the order of summation (i.e. replace $k$ by $r+1-k$) to get

$${\int }_1^{n+1}{b}_r(x)=\frac{1}{r+1}\sum _{k=0}^r\left(\genfrac{}{}{0pt}{}{r+1}{r+1-k}\right)B_k{(n+1)}^{r+1-k}=\frac{1}{r+1}\sum _{k=0}^r\left(\genfrac{}{}{0pt}{}{r+1}{k}\right)B_r{(n+1)}^{r+1-k}$$

which is equal to ${\sum }_{k=0}^n{k}^r$ (see the parent (http://planetmath.org/SumOfKthPowersOfTheFirstNPositiveIntegers) article).