Closed-form expression for $\sum_{k=0}^n\binom{n}kk^p$ for integers $n,\,p$ Is there a closed-form expression for the sum $\sum_{k=0}^n\binom{n}kk^p$ given positive integers $n,\,p$? Earlier I thought of this series but failed to figure out a closed-form expression in $n,\,p$ (other than the trivial case $p=0$).
$$p=0\colon\,\sum_{k=0}^n\binom{n}kk^0=2^n$$
I know that $\sum_{k=0}^n\binom{n}k=2^n$ and $\sum_{k=0}^nk^n=\frac{k^{n+1}-1}{k-1}$ but I am unsure of whether these would be of much use now.

Additionally, what about the similar series $\sum_{k=0}^n\binom{n}kk^n$ where $p=n$? 
 A: Let 
$$
f(x)=(e^x+1)^n=\sum_{k=0}^n \binom{n}{k}e^{kx}.
$$
Then 
$$
\left(\frac{d}{dx}\right)^p f(x)=\sum_{k=0}^n\binom{n}{k}k^pe^{kx}.$$
Plug in $x=0$. 
A: We know that $(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k$.Differentiate this, to get $n(1+x)^{n-1}=\sum_{k=0}^n \binom{n}{k} k x^{k-1}$. Multiply by $x$ to get $nx(1+x)^{n-1}=\sum_{k=0}^n \binom{n}{k} k x^k$. Take $x=1$ to get the first sum, And repeat this process for sums involving higher powers of $k$.
A: Use the identity $k\dbinom{n}k=n\dbinom{n-1}{k-1}$: for $p=1$ you get
$$\sum_k\binom{n}kk=n\sum_k\binom{n-1}{k-1}=n\sum_k\binom{n-1}k=n2^{n-1}\;.$$
For $p=2$:
$$\begin{align*}
\sum_k\binom{n}kk^2&=n\sum_k\binom{n-1}{k-1}k\\
&=n\sum_k\binom{n-1}k(k+1)\\
&=n\sum_k\binom{n-1}kk+n\sum_k\binom{n-1}k\\
&=n(n-1)\sum_k\binom{n-2}{k-1}+n2^{n-1}\\
&=n(n-1)\sum_k\binom{n-2}k+n2^{n-1}\\
&=n(n-1)2^{n-2}+n2^{n-1}\;.
\end{align*}$$
If you carry out the same sort of computation with $p=3$, you get
$$\sum_k\binom{n}kk^3=n(n-1)(n-2)2^{n-3}+2n(n-1)2^{n-2}+n2^{n-1}\;,$$
which can be written with falling factorials as $$\sum_k\binom{n}kk^3=n^{\underline3}2^{n-3}+3n^{\underline2}2^{n-2}+n^{\underline1}2^{n-1}\;.$$
After a little more experimentation one may conjecture and prove by induction that
$$\sum_k\binom{n}kk^p=\sum_{k=1}^p{p\brace k}n^{\underline k}2^{n-k}\;,$$
where $p\brace k$ is the Stirling number of the second kind.
A: A relate problem. Try this formula
$$ \sum_{k=0}^n\binom{n}kk^p= 2^n\sum_{k=0}^{p}\begin{Bmatrix} p\\k \end{Bmatrix} {n\choose k}2^{-k}k!, $$
where $p \in \mathbb{N}$ and $\begin{Bmatrix} p\\k \end{Bmatrix}$ is the Stirling numbers of the second kind. You can plug in $p=n$ in the above formula. 
A: If you consider $p$ as fixed, then the below can be considered as closed form I suppose:
$$\sum_{k=0}^{n} \binom{n}{k} k^p = \sum_{k=1}^{p} S(p,k) n(n-1)\dots(n-k+1) 2^{n-k} \quad \quad (1)$$
where $S(k,p)$ is a Stirling number of the second kind.
If you denote the operator of differentiating and multiplying by $x$ as $D_{x}$
Then we have that
$$(D_{x})^{n}f(x) = \sum_{k=1}^{n} S(n,k) f^{k}(x) x^{k}$$
where $S(n,k)$ is the Stirling number of the second kind and $f^k(x)$ is the $k^{th}$ derivative of $f(x)$.
This can easily be proven using the identity $$S(n,k) = S(n-1,k-1) + k \cdot S(n-1,k)$$
To prove $(1)$ above, we apply $D_x$, $p$ times to $(1+x)^n$, and set $x=1$.
A: Suppose we seek to evaluate
$$\sum_{k=0}^n {n\choose k} k^p.$$
Introduce
$$k^p = \frac{p!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{p+1}} \exp(kz) \; dz.$$
This yields for the sum
$$\frac{p!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{p+1}} 
\sum_{k=0}^n {n\choose k} \exp(kz) \; dz
\\ = \frac{p!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{p+1}} 
(1+\exp(z))^n \; dz
\\ = \frac{p!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{p+1}} 
(2+\exp(z)-1)^n \; dz
\\ = \frac{p!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{p+1}} 
\sum_{q=0}^n {n\choose q} (\exp(z)-1)^q 2^{n-q} \; dz
\\ = \sum_{q=0}^n {n\choose q} \times q! \times  2^{n-q} \times
\frac{p!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{p+1}} 
\frac{(\exp(z)-1)^q}{q!} \; dz.$$
Recall the species equation for labelled set partitions:
$$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which yields the bivariate generating function of the Stirling numbers
of the second kind
$$\exp(u(\exp(z)-1)).$$
Substitute this into the sum to get
$$\sum_{q=0}^n {n\choose q} \times q! \times  2^{n-q}
\times {p\brace q}$$
Now observe  that when $n\gt  p$ the Stirling  number is zero  for the
values $p\lt q \le n$ so we  may replace $n$ by $p.$ On the other hand
when $n\lt p$ the binomial coefficient  is zero for the values $n\lt q
\le p$ so we may again replace $n$ by $p.$ This finally yields
$$\sum_{q=0}^p {n\choose q} \times q! \times  2^{n-q}
\times {p\brace q}$$
as observed by the other contributors.
A: We have:
$$\sum_{k=0}^n {n\choose k} k^p =
\sum_{k=0}^n {n\choose k} \left[\sum_{j=0}^p j! {k\choose j} {p\brace j}\right] =
\sum_{j=0}^p j! \left[\sum_{k=j}^n {n\choose k} {k\choose j} \right] {p\brace j} =
\sum_{j=0}^p j! 2^{n-j}{n\choose j} {p\brace j} ,$$
with $p\brace j$ being the Stirling number of the second kind.
