In a question i wrote here, i pondered how to calculate the following sum;
$$\sum_{i=1}^m{(m+1-i)i^p}=m\cdot 1^p+(m-1)\cdot 2^p+...+2\cdot (m-1)^p+1\cdot m^p$$
Rearranging this gives me
$$\sum_{i=1}^1i^p+\sum_{i=1}^2i^p+...+\sum_{i=1}^{m-1}i^p+\sum_{i=1}^mi^p=\sum_{k=1}^m\sum_{i=1}^ki^p$$
Now, applying Faulhaber's Formula, i get
\begin{eqnarray*}\sum_{k=1}^m\sum_{i=1}^ki^p&=&\frac{1}{p+1}\sum_{k=1}^m\sum_{j=0}^p{\binom{p+1}{j}B_jk^{p+1-j}}\\&=&\frac{1}{p+1}\sum_{j=0}^p{\binom{p+1}{j}}B_j\sum_{k=1}^mk^{p+1-j}\end{eqnarray*}
Now i think I can apply Faulhaber's Formula again on the sum to the right to get
\begin{eqnarray*}\sum_{k=1}^m\sum_{i=1}^ki^p&=&\frac{1}{p+1}\sum_{j=0}^p{\binom{p+1}{j}}B_j\frac{1}{p+2-j}\sum_{i=0}^{p+1-j}\binom{p+2-j}{i}B_im^{p+2-j-i}\\&=&\sum_{j=0}^{p}\sum_{i=0}^{p+1-j}\binom{p+1}{j}\binom{p+2-j}{i}\frac{B_iB_j}{(p+1)(p+2-j)}m^{p+2-i-j}\\&-&\sum_{j=0}^{p}\sum_{i=0}^{p+1-j}\frac{B_iB_jp!}{i!j!(p+2-i-j)!}m^{p+2-i-j}\end{eqnarray*}
Is it possible to write this as a single sum or reduce this equation any further?