What does $\sum_{r=1}^{n} r^{k}$ equal in general I know that  $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$,  $\sum_{r=1}^{n} r^{2} = \frac{n(n+1)(2n+1)}{6}$, $ \sum_{r=1}^{n} r^{3} = \frac{n^2(n+1)^2}{4}$
but what does  $\sum_{r=1}^{n} r^{k}$ equal in general ?
Can we express it in some simple predetermined form

We may also express this in terms of Stirling numbers of the second
kind and falling factorials. We start with
$$\sum_{r=0}^n r^k
= k! [z^k] \sum_{r=0}^n \exp(rz)
= k! [z^k] \frac{\exp((n+1)z)-1}{\exp(z)-1}
\\ = k! [z^k] \frac{1}{\exp(z)-1} 
\sum_{q=1}^{n+1} {n+1\choose q} (\exp(z)-1)^q
\\ = k! [z^k]
\sum_{q=1}^{n+1} {n+1\choose q} (\exp(z)-1)^{q-1}
\\ = k! [z^k]
\sum_{q=1}^{n+1} (n+1)^{\underline{q}} 
\frac{1}{q}  \frac{(\exp(z)-1)^{q-1}}{(q-1)!}
\\ = \sum_{q=1}^{n+1} (n+1)^{\underline{q}} 
\frac{1}{q} {k\brace q-1}.$$
Note that we may set the upper limit of the sum to $k+1.$  If $n+1\gt
k+1$ we may lower to $k+1$ because the removed terms from the range
$k+2\le q\le  n+1$ produce zero by the Stirling number. If $k+1\gt n+1$
we may raise to $k+1$ because the extra terms from the range $n+2\le q\le
k+1$ produce zero through the  falling factorial.
We get
$$\sum_{q=2}^{k+1} (n+1)^{\underline{q}} 
\frac{1}{q} {k\brace q-1}
= (n+1) \sum_{q=2}^{k+1} n^{\underline{q-1}} 
\frac{1}{q} {k\brace q-1}$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{
\sum_{r=0}^n r^k =
(n+1)
\sum_{q=1}^{k} n^{\underline{q}} 
\frac{1}{q+1} {k\brace q}.}$$
In this way we get e.g. with $k=4$
$$(n+1) \times
\left[
n^{\underline{1}} \frac{1}{2} {4\brace 1} +
n^{\underline{2}} \frac{1}{3} {4\brace 2} +
n^{\underline{3}} \frac{1}{4} {4\brace 3} +
n^{\underline{4}} \frac{1}{5} {4\brace 4}
\right]$$
The Stirling numbers may be evaluated by inspection:
$$(n+1) \times
\left[
n^{\underline{1}} \frac{1}{2} 
\times 1 +
n^{\underline{2}} \frac{1}{3} 
\times \left( \frac{1}{2} {4\choose 2} + {4\choose 1} \right) +
n^{\underline{3}} \frac{1}{4} 
\times {4\choose 2} +
n^{\underline{4}} \frac{1}{5}
\times 1
\right]$$
We find
$$\sum_{r=0}^n r^4 =
(n+1) \times
\left[
\frac{1}{2} n^{\underline{1}} +
\frac{7}{3} n^{\underline{2}} +
\frac{3}{2} n^{\underline{3}} +
\frac{1}{5} n^{\underline{4}}
\right].$$
