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
$$\sum_{r=0}^n r^4 = (n+1) \times \left[ \frac{1}{2} n^{\underline{1}} + \frac{7}{3} n^{\underline{2}} + \frac{3}{2} n^{\underline{3}} + \frac{1}{5} n^{\underline{4}} \right].$$
