Summation of combination of binomial coefficient Is there any way to find:

$$\sum_{i=0}^n {\binom{n}{i}i^k}$$

I know that we can find it for small k by using binomial theorem by differentiating both sides and then multiplying both sides by x and repeating till the form presents itself and then putting x =1, but is there a way for any general power( k here) of i?
I think that the general form of $$S_k=\sum_{i=0}^n {\binom{n}{i}i^k}$$ is "just" an hypergeometric function
$$S_k=n \, _kF_{k-1}(2,\cdots,2,1-n;1,\cdots,1;-1)$$ in which the arguments appear $(k-1)$ times.

Looking for cases, the result seems to be of the form
$$S_k=2^{n-k}\,n \, P_{k-1}(n)$$ where $P$ is a polynomial.

For the very first values of $k$, the results are
$$\left(
\begin{array}{cc}
 k &amp; P_{k-1}(n) \\
 1 &amp; 1 \\
 2 &amp; n+1 \\
 3 &amp; n^2+3 n \\
 4 &amp; n^3+6 n^2+3 n-2 \\
 5 &amp; n^4+10 n^3+15 n^2-10 n \\
 6 &amp; n^5+15 n^4+45 n^3-15 n^2-30 n+16 \\
 7 &amp; n^6+21 n^5+105 n^4+35 n^3-210 n^2+112 n \\
 8 &amp; n^7+28 n^6+210 n^5+280 n^4-735 n^3+28 n^2+588 n-272 \\
 9 &amp; n^8+36 n^7+378 n^6+1008 n^5-1575 n^4-2436 n^3+5292 n^2-2448 n n+7936
\end{array}
\right)$$ where 


the second coefficient is $\binom{k+1}{2}=\frac{1}{2} k (k+1)$
the third coefficient is $\binom{\binom{k}{2}}{2}=\frac{1}{8} (k-2) (k-1) k (k+1)$
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\begin{align}
\sum_{i = 0}^{n}{n \choose i}i^{k} &amp; =
\sum_{i = 0}^{n}{n \choose i}\bracks{z^{k}}\pars{k!\expo{iz}} =
k!\bracks{z^{k}}\sum_{i = 0}^{n}{n \choose i}\pars{\expo{z}}^{i} \\[5mm] &amp; =
\bbx{k!\bracks{z^{k}}\pars{1 + \expo{z}}^{n}}
\end{align}