Generating function for $\sum_i x^i i^n$ I am trying to find a generating function for the following family of infinite sums:
$$F_n(x)= \sum_{i \geq 0} x^i i^n$$
Solving for small $n$ is trivially done by hand. I entered the sum into WolframAlpha for small $n$, and noticed that it always resolves to $p(x) / (1-x)^{n+1}$ where $p(x)$ is always a nice looking polynomial (order $n$, small positive integer coefficients, etc.).
Is there a closed form for $p$ or its coefficients?
 A: By classifying the functions from $\{1,2,\ldots,n\}$ to $\{1,2,\ldots,i\}$ according to the cardinality of their range we have that
$$ i^n = \sum_{k=0}^{n}\binom{i}{k}k!{n\brace k} $$
where ${n\brace k}$ is a Stirling number of the second kind, representing in how many ways we may partition $\{1,2,\ldots,n\}$ into $k$ non-empty subsets. This allows to state that
$$ \sum_{i\geq 0} i^n x^i = \sum_{k=0}^{n}k!{n\brace k}\sum_{i\geq 0}\binom{i}{k}x^i = \sum_{k=0}^{n}k!{n\brace k}\frac{x^k}{(1-x)^{k+1}}$$
where the last identity follows from stars and bars. We may also notice that the claim
$$ (1-x)^{n+1}\sum_{i\geq 0}i^n x^i\text{ is a polynomial } $$
is a consequence of a well-known fact. By defining the forward difference operator $\delta$ through $(\delta p)(x) = p(x+1)-p(x)$, we have that if $p(x)$ has degree $d\geq 1$, then $(\delta p)(x)$ has degree $d-1$. In particular, by applying $\delta^{n+1}$ to a polynomial with degree $n$ we always get $0$. This is just an equivalent form of the previous statement, since it proves
$$ \forall m>n,\qquad [x^m]\left((1-x)^{n+1}\sum_{i\geq 0}i^n x^i\right) = 0. $$
