Idea for recursively generating formulas for sums of powers of integers. Is it correct? I am not even a beginner mathematician, being unable to understand simple proofs others easily grasp. Still I noodle around and I think I stumbled on a fun way to determine the formula for sums of powers of integers, i.e. $f(n) = 1^p + 2^p + 3^p + \cdots +n^p$, based on knowing the formulas for the previous powers $1, \ldots, p-1$.
Let $\operatorname{Sum}(g(x))$ mean the sum from $1$ to $n$ of $g(x)$. Then, for example, one can derive the formula for $\operatorname{Sum}(x^4)$ by combining the formulas for $\operatorname{Sum}(x)$, $\operatorname{Sum}(x^2)$, $\operatorname{Sum}(x^3)$ in a special way.
How so? Consider
$$\begin{align}
\operatorname{Sum}((x+1)^{p+1} - (x^{p+1})) &= 2^{p+1} - 1^{p+1} + 3^{p+1} - 2^{p+1} + ..(n+1)^{p+1} - n^{p+1} \\
&= (n+1)^{p+1} - 1^{p+1}
\end{align}$$ because all the interior terms cancel out.
But also, using the binomial expansion,
$$\begin{align}
\operatorname{Sum}((x+1)^{p+1} - (x^{p+1})) &= \operatorname{Sum}(c_1x^p + c_2x^{p-1} + c_3x^{p-2} + \cdots + 1) \\
&= c_1\operatorname{Sum}(x^p) + c_2\operatorname{Sum}(x^{p-1}) + \cdots +\operatorname{Sum}(1) \\
&= (n+1)^{p+1} - 1
\end{align}$$
so that
$$c_1\operatorname{Sum}(x^p) = (n+1)^{p+1} - 1 - \left(c_2\operatorname{Sum}(x^{p-1}) + c_3\operatorname{Sum}(x^{p-2}) +  \cdots + \operatorname{Sum}(1)\right) \tag{1}$$
Since we stipulated that we know the formulas for $\operatorname{Sum}(x^m)$ from $1, \ldots, p-1$, and since all the $c_k$ are known, as well as $(n+1)^{p+1}-1$, we can derive $\operatorname{Sum}(x^p)$ with a little algebra.
To see this in action, let's derive the formula for the sum of squares, i.e. $p=2$. From $(1)$ we have
$$\begin{align}
3\operatorname{Sum}(x^2) &= (n+1)^3 - 1 - \left(3\operatorname{Sum}(x^1) + \operatorname{Sum}(1)\right) \\
&= (n+1)^3 - 1 - 3n(n+1)/2 -n
\end{align}$$
since $\operatorname{Sum}(x) = n(n+1)/2$ and $\operatorname{Sum}(1) = n$
The above simplifies to $\operatorname{Sum}(x^2) = n(n+1)(2n+1)/6$,  agreeing with the known formula.
To show this is not a fluke, let's derive the formula for $\operatorname{Sum}(x^3)$. By $(1)$ we have
$$\begin{align}
4\operatorname{Sum}(x^3) &= (n+1)^{4} - 1 - \left(6\operatorname{Sum}(x^2) + 4\operatorname{Sum}(x) + \operatorname{Sum}(1)\right) \\
&= (n+1)^4 - 1 - \left(n(n+1)(2n+1) + 2n(n+1) + n\right) \\
&= n^2(n+1)^2
\end{align}$$
Thus $\operatorname{Sum}(x^3) = n^2(n+1)^2/4$, which agrees with the known formula.
Undoubtedly this is a well known approach, and I claim absolutely zero originality.

I've been informed by someone that this is a well-known result and not value added. I apologize to the community.
