(I wanted to comment on Ira Gessel's answer, but can't - therefore as a standalone answer)
Have a look at the excellent and comprehensive article Eulerian Polynomials: from Euler's Time to the Present from Dominique Foata.
In it you can find the explicit formula (page 12, (2.8)) $$\sum_{i=1}^{m}{i^nt^i}=\sum_{l=1}^{n}{(-1)^{n+l}{{n}\choose{l}}\frac{t^{m+1}A_{n-l}(t)}{(t-1)^{n-l+1}}}+(-1)^n\frac{t(t^m-1)}{(t-1)^{n+1}}A_{n}(t)$$ where $A_n$ is the usual Eulerian polynomial.
This formula is in essence what Ira Gessel has described already.