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?
 A: 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
$$\left(
\begin{array}{cc}
 k & P_{k-1}(n) \\
 1 & 1 \\
 2 & n+1 \\
 3 & n^2+3 n \\
 4 & n^3+6 n^2+3 n-2 \\
 5 & n^4+10 n^3+15 n^2-10 n \\
 6 & n^5+15 n^4+45 n^3-15 n^2-30 n+16 \\
 7 & n^6+21 n^5+105 n^4+35 n^3-210 n^2+112 n \\
 8 & n^7+28 n^6+210 n^5+280 n^4-735 n^3+28 n^2+588 n-272 \\
 9 & n^8+36 n^7+378 n^6+1008 n^5-1575 n^4-2436 n^3+5292 n^2-2448 n n+7936
\end{array}
\right)$$ where 


*

*the second coefficient is $\binom{k+1}{2}=\frac{1}{2} k (k+1)$

*the third coefficient is $\binom{\binom{k}{2}}{2}=\frac{1}{8} (k-2) (k-1) k (k+1)$
A: I am not aware of a closed formula for such numbers but they satisfy 
interesting identities and relations with other numbers, besides those already pointed out in the precedent answers.
First point is that we can omit the sum limits, because they are implicit in the binomial coefficient.
So we can write (actual sum limits are indicated in brackets)
$$
s(n,k) = \sum\limits_{i = 0}^n {\binom{n}{i}i^{\,k} } 
 = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {\binom{n}{i}i^{\,k} } 
$$
In this way some of the algebraic manipulations are much simplified.
Interesting is the recurrence in $k$
$$
\eqalign{
  & s(n,k + 1) = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {\binom{n}{i}i^{\,k + 1} }
  = \sum\limits_{\left( {0\, \le \,1\, \le } \right)\,i\,\left( { \le \,n} \right)} {i\binom{n}{i}i^{\,k} }  =   \cr 
  &  = n\sum\limits_{\left( {\,1\, \le } \right)\,i\,\left( { \le \,n} \right)} {\binom{n-1}{i-1}i^{\,k} }
    = n\sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n - 1} \right)} {\binom{n-1}{i}\left( {i + 1} \right)^{\,k} }  =   \cr
  &  = n\sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,k} \right)} { \binom{k}{j}
\sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n - 1} \right)} { \binom{n-1}{i} i^{\,j} } }  =   \cr 
  &  = n\sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,k} \right)} {  \binom{k}{j} s(n - 1,j)}  \cr} 
$$
with the initial values
$$
\eqalign{
  & s(n,0) = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {\binom{n}{i}i^{\,0} }
  = 2^n  \cr 
  & s(0,k) = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {\binom{0}{i}i^{\,k} } 
 = 0^{\,k}  = \left[ {0 = k} \right] \cr} 
$$
where $[P]$ denotes the Iverson bracket.
The recurrence resembles that for Bernoulli numbers/polynomials 
(except for the multiplicand in $n$) .
Another interesting relation is that with the Stirling numbers of 2nd kind
$$
\eqalign{
  & s(n,k) = \sum\limits_{i = 0}^n {
 \left( \matrix{ n \cr  i \cr}  \right)i^{\,k} } 
  = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {
  \left( \matrix{n \cr   i \cr}  \right)i^{\,k} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {
  \left( \matrix{n \cr   i \cr}  \right)
\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {
 \left\{ \matrix{ k \hfill \cr   j \hfill \cr}  \right\}i^{\,\underline {\,j\,} } } }  = 
  \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {
 \left\{ \matrix{k \hfill \cr  j \hfill \cr}  \right\}\sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {
 \left( \matrix{ n \cr   i \cr}  \right)i^{\,\underline {\,j\,} } } }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {
 j!\left\{ \matrix{  k \hfill \cr   j \hfill \cr}  \right\}
\sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {
 \left( \matrix{ n \cr   i \cr}  \right)\left( \matrix{  i \cr   j \cr}  \right)} }
  = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {
  j!\left\{ \matrix{k \hfill \cr   j \hfill \cr}  \right\}\left( \matrix{  n \cr   j \cr}  \right)2^{n - j} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {
  \left\{ \matrix{  k \hfill \cr  j \hfill \cr}  \right\}n^{\,\underline {\,j\,} } 2^{n - j} }  \cr} 
$$
where $n^{\,\underline {\,j\,} }$represents the Falling Factorial.
Also interesting is that by the inversion of the binomial convolution we get
$$
s(n,k) = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {
  \binom{n}{i}i^{\,k} } \quad  \Leftrightarrow \quad n^{\,k}
  = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n} \right)} {
\left( { - 1} \right)^{\,n - i}  \binom{n}{i}s(i,k)} 
$$
Finally - as an integration to @Felix Marin's answer which is too long
for a comment - note that
$$
\eqalign{
  & \left( {e^{\,x}  + 1} \right)^n  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
 \left( \matrix{ n \hfill \cr   k \hfill \cr}  \right)e^{\,k\,x} }
  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\sum\limits_{0\, \le \,j} {
  \left( \matrix{n \hfill \cr   k \hfill \cr}  \right){{\,k^{\,j} \,x^{\,j} } \over {j!}}} }   \cr 
  & \sum\limits_{0\, \le \,k} {
 \left( \matrix{ n \hfill \cr   k \hfill \cr}  \right)k^{\,m} }
  = \left. {{{d^{\,m} } \over {dx^{\,m} }}\left( {e^{\,x}  + 1} \right)^n } \right|_{\,x = 0}   \cr 
  & \sum\limits_{0\, \le \,k} {s(n,k){{\,\,x^{\,k} } \over {k!}}}  = \left( {e^{\,x}  + 1} \right)^n  \cr} 
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{i = 0}^{n}{n \choose i}i^{k} & =
\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] & =
\bbx{k!\bracks{z^{k}}\pars{1 + \expo{z}}^{n}}
\end{align}
