For intuition, think about converting your sum to an integral. $\sum_{i=0}^n i^r \approx \int_0^n i^r di=\frac 1{r+1}n^{r+1}$
If you try it for small values, for $r=0$ you get the sum of $n \ 1$'s, which is $n$ (growing like $r+1=1$). For $r=1$ you get the triangular numbers $\frac {n(n+1)}2$, (growing like $r+1=2$).