Original question is a duplicate of this question. Please refer to after the edit
Original question:
I was investigating formulas of the form $\sum_{k=0}^n k^a \binom{n}{k}$ after noticing on Wikipedia that $\sum_{k=0}^n k \binom{n}{k} = n2^{n-1}$ and $\sum_{k=0}^{n} k^2 \binom{n}{k} = (n+n^2)2^{n-2}$.
Wolfram Alpha notes that for $a=3$, $\sum_{k=0}^n k^3 \binom{n}{k} = n^2 (n + 3)2^{n-3}$ but for $a=4,5$ then switches to $\sum_{k=0}^n k^4 \binom{n}{k} = n\ _4 F_3(2, 2, 2, 1 - n;1, 1, 1;-1)$ and $\sum_{k=0}^n k^5 \binom{n}{k} = n\ _5 F_4(2, 2, 2, 2, 1 - n;1, 1, 1, 1;-1)$.
My conjecture is that, in general, $$\sum_{k=0}^n k^a \binom{n}{k} = n\ _a F_{a-1}(2, 2, ..., 2, 1 - n;1, 1, ..., 1; -1)$$ Which is some kind of polynomial since one of the numerators there is non-positive (and I suspect Wolfram Alpha just spat this function out so as to avoid calculating the polynomial outright), but I don't know how to use this function for computing with a given $n$ and $a$ (though I begin to suspect there isn't an easy way to do so).
I'm also wondering if there's a formula at all for $\sum_{k=0}^n k^n \binom{n}{k}$.
Any suggestions or advice welcome.
EDIT
Since my original question already has an answer, I'd like to pivot to asking how the resulting identity works. $$\begin{align} \newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}} \sum_{k=0}^n k^a \binom{n}{k} &= n\ _a F_{a-1}(2, 2, ..., 2, 1 - n;1, 1, ..., 1; -1)\\ &=2^{n-m}\sum_{j=0}^m\binom{n}{j}2^{m-j}\stirtwo{m}{j}j!\\ \end{align}$$
$$S_n=\sum_{k=0}^n k^n \binom{n}{k}$$ is sequence $A072034$ in $OEIS$
Vaclav Kotesovec proposed a superb approximation $$S_n \sim \frac{1}{\sqrt{1+W\left(\frac{1}{e}\right)}}\left(\frac{n}{e\, W\left(\frac{1}{e}\right)}\right)^n$$ where $W(.)$ is Lambert function.