Sum of Binomial Coefficients Times a Polynomial Is there a closed for expression for,
$\displaystyle\sum_{k=0}^n {n\choose k}k^2$
It holds that,
$\displaystyle\sum_{k=0}^n {n\choose k}k=n2^{n-1}$
Is there a generalization for higher degrees?
 A: For a different solution for a general case that the given by @JackD'Aurizio you want to transform the polynomial $p(k)$ in the expression
$$\sum_k\binom{n}{k}p(k)$$
in a polynomial based in falling factorials. By example
$$\sum_k\binom{n}{k}k^2=\sum_k\binom{n}{k}(k^{\underline{2}}+k)$$
Then the binomial sum can be simplified extracting falling factorials of $n$ out of the summation and (probably) changing the index of the sums.
To pass from $p(k)$ to a "polynomial" of falling factorials you will need to know that
$$k^n=\sum_j \left\{{ n\atop j }\right\}k^{\underline j}$$
for $n\ge 0$, where the symbols $\left\{{ n\atop j }\right\}$ are Stirling numbers of the second kind.
This link have an easy algorithm to pass an entire polynomial to a "polynomial" of falling factorials.
A: For the factor $k$, 
$$\sum_{k=1}^n {n\choose k}k=\sum_{k=1}^n \frac{n!}{(n-k)!k!}k
=n\sum_{k=1}^n \frac{(n-1)!}{(n-k)!(k-1)!}\\
=n\sum_{k=0}^{n-1} {n\choose k}=n2^{n-1}.$$
Similarly for $(k-1)k$,
$$\sum_{k=2}^n {n\choose k}(k-1)k=\sum_{k=1}^n \frac{n!}{(n-k)!k!}(k-1)k=(n-1)n\sum_{k=1}^n \frac{(n-2)!}{(n-k)!(k-2)!}\\
=(n-1)n\sum_{k=0}^{n-2} {n\choose k}=(n-1)n2^{n-2}.$$
More generally,
$$\sum_{k=m}^n {n\choose k}(k-m+1)\cdots(k-1)k\\
=\sum_{k=1}^n \frac{n!}{(n-k)!k!}(k-m+1)\cdots(k-1)k\\
=(n-m+1)\cdots(n-1)n\sum_{k=1}^n \frac{(n-m)!}{(n-k)!(k-m)!}\\
=(n-1)n\sum_{k=0}^{n-m} {n\choose k}=(n-m+1)(n-1)n2^{n-m}.$$
Then,
$$k^2=(k-1)k+k,\\
k^3=(k-2)(k-1)k+3k^2-2k=(k-2)(k-1)k+3(k-1)k+k\\
\cdots$$
and you can convert any polynomial to a sum of falling factorials.
Notice that the $m$ first terms in the summations must be added separately.
