On sums of powers of natural numbers

The problem of finding the sum of a polynomial’s values is considered. In particular, for any $n\ge 3$, the explicit formula for the sum of the $n$th powers of natural numbers $S_n={\sum }_{x=1}^m{x}^n$ is proved: $$\sum _{x=1}^m{x}^n=(-1{)}^{n}m(m+1)(-\frac{1}{2}+\sum _{i=2}^n{a}_i(m+2)(m+3)\mathrm{.}\mathrm{.}\mathrm{.}(m+i)),$$ here $a_i=\frac{1}{i+1}{\sum }_{k=1}^i\frac{(-1{)}^k{k}^n}{k!(i-k)!}$, $(i=2,3,\mathrm{.}\mathrm{.}\mathrm{.},n-1)$, $a_n=\frac{(-1{)}^n}{n+1}$. Note that this formula does not contain Bernoulli numbers.