Simplification of Faulhaber's formula

Question

The fact that $$\displaystyle\sum_{k=1}^{n-1}k^m=\dfrac{1}{m+1}\displaystyle\sum_{s=0}^m \dbinom{m+1}{s}\mathrm B_{s}n^{m-s+1} \tag*{(1)}$$ should imply $$\displaystyle\sum_{k=1}^n k^m=\dfrac{1}{m+1}\displaystyle\sum_{s=0}^m (-1)^s \dbinom{m+1}{s}\mathrm B_{s}n^{m-s+1}. \tag*{(2)}$$ It should be true because $\mathrm B_{2s+1}=0$ for $s\ge 1$, but I don't understand how $\mathrm B_{2s+1}=0$ implies $(2)$. If I change every $n$ to $n+1$ in $(1)$, then $$\begin{align*}\displaystyle\sum_{k=1}^n k^m&=\dfrac{1}{m+1}\displaystyle\sum_{s=0}^m \dbinom{m+1}{s}\mathrm B_{s}\displaystyle\sum_{t=0}^{m-s+1}\dbinom{m-s+1}{t}n^t\\&=\dfrac{1}{m+1}\displaystyle\sum_{s=0}^m\displaystyle\sum_{t=0}^{m-s+1}\mathrm B_{s}\dbinom{m+1}{s}\dbinom{m-s+1}{t}n^t\\&=\dfrac{1}{m+1}\displaystyle\sum_{s=0}^m\displaystyle\sum_{t=0}^{m-s+1}\mathrm B_{s}\dbinom{m+1}{s+t}\dbinom{s+t}{s}n^t\\&=\dfrac{1}{m+1}\displaystyle\sum_{s=0}^m\displaystyle\sum_{t=0}^{m-s+1}\dfrac{\mathrm B_{s}}{s! \, t! \, (s+t)!}\dbinom{m+1}{s+t}n^t,\end{align*}$$ but I don't know how to proceed. Is there way to get rid of the double sum?

Answer 1

The summands in (1) with $s$ even are unaffected by multiplication by $(-1)^s=1$. The terms with $s$ odd and greater than 1 are also unaffected because all such terms are 0, thanks to the factor of $B_s=0$. Thus, the only term that is affected is the term with $s=1$, and it is easy to check that this change on the right hand side changes the sum in precisely the same way as the change on the left hand side.