# $\mathbb{Z}$-Polynomials in an Enumeration Identity

I've conjectured the following identity: For $1 \leqslant k \leqslant l \leqslant n$ and $m \in \mathbb{N}$, \begin{align} \sum_{1 \leqslant i_1 < \cdots < i_l \leqslant n} i_{k}^{m} = \sum_{j = 1}^{m} P_{m,j}(k) \binom{n+j}{l+m}, \end{align} where $P_{m,j}(k)$ generates the $\mathbb{Z}$-polynomial triangle: \begin{align} \begin{array}{cccccc} m / j & 1 & 2 & 3 & 4 & 5 \\ 1 & k & & & & \\ 2 & k & k^{2} & & & \\ 3 & k & 3k^{2} + k & k^{3} & & \\ 4 & k & 7k^{2} + 4k & 6 k^{3} + 4k^{2} + k & k^{4} & \\ 5 & k & 15 k^{2} + 11 k & 25 k^{3} + 30 k^{2} + 11k & 10 k^{4} + 10 k^{3} + 5 k^{2} + k & k^{5} \end{array} \quad etc \end{align} In particular, \begin{align} P_{m,1}(k) & = k \\ P_{m,2}(k) & = (2^{m-1} - 1) k^{2} + (2^{m-1} - m) k \\ P_{m,m-1}(k) & = \sum_{j = 1}^{m-1} \binom{m}{j+1} k^{m-j} \\ P_{m,m}(k) & = k^{m}. \end{align} When $k = 1$, the polynomials specialize to Eulerian numbers. Summing over $j$, I conjecture \begin{align} \sum_{j = 1}^{m} P_{m,j}(k) = \sum_{l = 0}^{m} |s(m,l)|k^{l} = (k)_{m}, \end{align} where $s(m,l)$ is the $(m,l)^{\text{th}}$-Stirling Number of the first kind. Are these polynomials well known?

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