we can write:
$\displaystyle\int_{0}^{m}{x^m}dx\le\sum_{n=1}^{m}{n^m}
\le\int_{0}^{m+1}{x^m}dx$

so $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k}{m+1}$ and then $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k_1}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1)-m(m+1)k_2}{m+1}$ in which assume that $k_2\le k$ and$k\le k_1$,


but since $(m+1)^{m+1}\equiv{m+1}\pmod{m}$ and also $m^{m+1}\equiv m\pmod{m}$ so the

 central statement should be zero, or $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1\bmod m$, when the first

 congruence holds for the numbers 1,2,6,42,1806, second one holds for these numbers.
 
note: if $m$ be an even number so $m^{m}-1$ is divided to $m+1$ then $(m+1)^{m+1}\equiv{m+1}\pmod{m(m+1)}$ and $m^{m+1}\equiv m\pmod{m(m+1)}$