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