I want to prove the following equality where $A(k,m)$ is the Eulerian number :
$$\forall k\ge0,\sum_{k=0}^{\infty}n^k x^k = \frac{\sum_{m=0}^{k-1}A(k,m)x^{m+1}}{(1-x)^{k+1}}$$
I previously proved that the two following identities are equivalent.
\begin{align*} \sum_{n\ge0}a_nt^n &= \frac{\sum_{j=0}^{k}\beta_jt^j}{(1-t)^{k+1}} \\ \forall n\ge0 , a_n &= \sum_{j=0}^{k}\beta_j\binom{n+k-j}{k} \end{align*}
It is advised to use this previous result to prove the equality. Since I only had a combinatorial definition of the Eulerian numbers, I managed to prove the closed-form expression : $A(n,m)=\sum_{k=0}^{m+1}(-1)^k \binom{n+1}{k} (m+1-k)^n.$ So by using the closed-form expression for Eulerian numbers and the previous equivalence I need to prove that :
$$\forall n\ge1 , n^k = \sum_{j=1}^{k} \sum_{i=0}^{j} (-1)^i\binom{k+1}{i}(j-i)^k\binom{n+k-j}{k}$$
But I have no idea how to go further, I don't see how it is an improvement to prove the original equality.
