TITLE: binomial expression of a powered term
QUESTION [5 upvotes]: One answer to a previous question of mine asserted that $$k^2=\binom k2+\binom {k+1}2.$$ 
I checked that the formula is true. However, it intrigued me. Is there a similar expression for $k^3$? How would I find a binomial for $k^n$? This is not a duplicate question to the best of my knowledge.

REPLY [2 votes]: I looked this up on OEIS A008292 and 
Wikipedia.
Apparently
$$k^m = \sum_{q=0}^{m-1} {k+q\choose m} 
\langle {m\atop q}\rangle$$
with  $\langle  {m\atop  q}\rangle$  the Eulerian  numbers  that  have
generating function
$$\frac{t-1}{t-\exp((t-1)z)}.$$
We now prove this summation formula.
We have the following exponential generating function
$$G_k(z) = \sum_{m\ge 0} k^m \frac{z^m}{m!} = \exp(kz).$$
On the other hand the sum formula gives the EGF
$$H_k(z) = \sum_{m\ge 0} 
\frac{z^m}{m!} \sum_{q=0}^{m-1} {k+q\choose m} 
\langle {m\atop q}\rangle.$$
We can extend this to $q=m$ because the Eulerian number 
$\langle {m\atop m}\rangle$ is zero to get
$$\sum_{m\ge 0} 
\frac{z^m}{m!} \sum_{q=0}^{m} {k+q\choose m} 
\langle {m\atop q}\rangle
\\ = \sum_{q\ge 0} \sum_{m\ge q} 
{k+q\choose m} \frac{z^m}{m!}
\langle {m\atop q}\rangle.$$
Now introduce
$${k+q\choose m}
= \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k+q}}{w^{m+1}} \; dw.$$
Substitute this into the sum to get
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\sum_{q\ge 0} \frac{(1+w)^{k+q}}{w}
\sum_{m\ge q} \frac{1}{w^m} \frac{z^m}{m!}
\langle {m\atop q}\rangle \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k}}{w}
\sum_{q\ge 0} (1+w)^q
\sum_{m\ge q} \frac{1}{w^m} \frac{z^m}{m!}
\langle {m\atop q}\rangle \; dw.$$
Now what we have here is a double
annihilated coefficient extractor
which we now collapse:
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k}}{w}
\sum_{q\ge 0} (1+w)^q
[t^q] \frac{t-1}{t-\exp((t-1)z/w)} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k}}{w}
\frac{w}{1+w-\exp(z)} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k}}{1+w-\exp(z)} \; dw.$$
The contribution  from the  pole at $w=\exp(z)-1$  which is  simple is
precisely
$$(1+(\exp(z)-1))^k = \exp(kz) = G_k(z),$$
QED.
There is another annihilated coefficient extractor at this
MSE link I and at this
MSE link II and also here at this MSE link III.
The Maple code for the initial lookup at the OEIS was as follows:

Q :=
proc(m)
    local s, sys, sol;

    s := expand(add(a[q]*binomial(k+q,m), q=0..m-1));

    sys := [coeff(s, k, m)=1];
    sys :=
    [op(sys), seq(coeff(s, k, q)=0, q=0..m-1)];

    sol := solve(sys, [seq(a[q], q=0..m-1)]);
    subs(sol[1], [seq(a[q], q=0..m-1)]);
end;