Closed form of $ \sum_{k_{j-1}=0}^{k_j}....\sum_{k_1=0}^{k_2} \sum_{k=0}^{k_1} k^m $ as a polynomial in $k_j$?

Question

Consider how the iterated sum $ \sum_{k_{j-1}=0}^{k_j}....\sum_{k_1=0}^{k_2} \sum_{k=0}^{k_1} 1 $ produces the diagonals on pascal's triangle, so it has a nice closed form. Does the iterated sum $ \sum_{k_{j-1}=0}^{k_j}....\sum_{k_1=0}^{k_2} \sum_{k=0}^{k_1} k^m $ have a nice formula as well? If $j=1$ we just have the Faulhaber formula for $m$, and if $j=2$ we have the sum over the Faulhaber polynomials etc... How can one calculate this without iteratively using Faulhaber's formula?

Answer 1

So, the sum can be rewritten as $$\sum_{0\leq k\leq k_1\leq\dots\leq k_j} k^m=\sum_{k=0}^{k_j} k^m \sum_{k\leq k_1\leq\dots\leq k_j} 1 = \sum_{k=0}^{k_j} k^m \binom{k_j-k+j-1}{j-1}$$ As explained in my other answer, this sum can be further reduced to $$\sum_{t=0}^m \left\langle m\atop t\right\rangle \binom{k_j+j+m-1-t}{j+m},$$ where the binomial coefficients represent polynomials in $k_j$ (for fixed $j$ and $m$).