general way to get formula as multiplication. We assume:
$$ 1^n + 2^n + 3^n + .. + k^n$$
where k and n are natural numbers.
Are there a general way to get it as multiplication?
For example:
$$ 1^3 + 2^3 + 3^3 + .. + k^3  = \binom {k+1} 2 ^2 $$

Suppose that for $0\le j\lt m$, there is a polynomial $P_j$, with degree $j+1$, so that
$$
\sum_{k=0}^nk^j=P_j(n)\tag{1}
$$
The binomial theorem says that
$$
(k+1)^{m+1}-k^{m+1}=\sum_{j=0}^m\binom{m+1}{j}k^j\tag{2}
$$
Sum $(2)$ in $k$ using $(1)$:
$$
\begin{align}
(n+1)^{m+1}
&=\sum_{k=0}^n\sum_{j=0}^m\binom{m+1}{j}k^j\\
&=\sum_{k=0}^n\left[(m+1)k^m+\sum_{j=0}^{m-1}\binom{m+1}{j}k^j\right]\\
&=(m+1)\sum_{k=0}^nk^m+\sum_{j=0}^{m-1}\binom{m+1}{j}P_j(n)\tag{3}
\end{align}
$$
Solving $(3)$ for $\sum\limits_{k=0}^nk^m$
$$
\begin{align}
\sum_{k=0}^nk^m
&=\frac1{m+1}\left[(n+1)^{m+1}-\sum_{j=0}^{m-1}\binom{m+1}{j}P_j(n)\right]\\
&=P_m(n)\tag{4}
\end{align}
$$
That is, $(1)$ is true for $j=m$.
Thus, we have shown that if $(1)$ is true for $0\le j\lt m$, then $(1)$ is also true for $j=m$. Since $P_0(n)=n+1$, we have that $(1)$ is true for $j=0$. Therefore, by induction, $(1)$ is true for all $j\ge0$.
Notice that $(4)$ also gives us a formula to compute $P_m$ from the $P_j$ for $j\lt m$.
$$
\begin{align}
P_1(n)
&=\frac12\left[\vphantom{\binom{2}{0}}\right.(n+1)^2-\binom{2}{0}\overbrace{(n+1)\vphantom{\frac{n^2+n}2}}^{P_0(n)}\left.\vphantom{\binom{2}{0}}\right]\\[4pt]
&=\frac{n^2+n}2\\
P_2(n)
&=\frac13\left[\vphantom{\binom{2}{0}}\right.(n+1)^3-\binom{3}{0}\overbrace{(n+1)\vphantom{\frac{n^2+n}2}}^{P_0(n)}-\binom{3}{1}\overbrace{\frac{n^2+n}2}^{P_1(n)}\left.\vphantom{\binom{2}{0}}\right]\\[4pt]
&=\frac{2n^3+3n^2+n}6\\
P_3(n)
&=\frac14\left[\vphantom{\binom{2}{0}}\right.(n+1)^4-\binom{4}{0}\overbrace{(n+1)\vphantom{\frac{n^2+n}2}}^{P_0(n)}-\binom{4}{1}\overbrace{\frac{n^2+n}2}^{P_1(n)}-\binom{4}{2}\overbrace{\frac{2n^3+3n^2+n}6}^{P_2(n)}\left.\vphantom{\binom{2}{0}}\right]\\[4pt]
&=\frac{n^4+2n^3+n^2}4
\end{align}
$$
