Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$?

Both sides are polynomials in $n$ of degree $8$. Since they coincide for $n=0,\dots,8$, they are equal.

Any $9$ points will do. Taking $n=-4,\dots,4$ is probably easier to do by hand.

The formula is already true for $n=1,2,....,5$ and we know that $$ \sum_{i=1}^ni= \frac{n\left(n+1\right)}{2}$$ Assume $$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4 =\frac{n^4\left(n+1\right)^4}{8}$$

then, $$\begin{align}\sum_{i=1}^{n+1} i^5+i^7&=\sum_{i=1}^{n} i^5+i^7 +(n+1)^5 +(n+1)^7\\&=\color{blue}{\frac{n^4\left(n+1\right)^4}{8}}+(n+1)^5 +(n+1)^7 \\&=\color{blue}{\frac{n^4\left(n+1\right)^4}{8}}+(n+1)^4\left[n+1 +(n+1)^3 \right] \\&=(n+1)^4\left( \frac{n^4}{8} +n+1 +\color{red}{n^3+3n^2+3n+1}\right) \\&=(n+1)^4\left( \frac{n^4}{8}+ n^3+3n^2+4n+2\right) \\&=\frac{(n+1)^4}{8}\left( n^4+ \color{blue}{4}\cdot\color{red}{2}\cdot n^3+\color{blue}{6}\cdot\color{red}{2^2}\cdot n^2+\color{blue}{4}\cdot\color{red}{2^3}\cdot n+\color{red}{2^4}\right) \\&=\color{blue}{\frac{\left(n+1\right)^4\left(n+2\right)^4}{8}=2\left( \sum_{i=1}^{n+1}i\right)^4}\end{align}$$

which prove that the formula is true

Notice $\sum_{k=1}^n k = \frac{n(n+1)}{2}$. For the identity at hand,

$$\sum_{k=1}^n k^5 + \sum_{k=1}^n k^7 \stackrel{?}{=} 2 \left(\sum_{k=1}^n k\right)^4$$ If one compute the difference of successive terms in RHS, we find

$$\begin{align}{\rm RHS}_n - {\rm RHS}_{n-1} &= 2\left(\frac{n(n+1)}{2}\right)^4-2\left(\frac{n(n-1)}{2}\right)^4\\ &= \frac{n^4}{8}\left((n+1)^4 - (n-1)^4\right) = \frac{n^4}{8}\left(8n^3 + 8n\right) = n^7 + n^5\end{align}$$ This clearly equals to ${\rm LHS}_n - {\rm LHS}_{n-1}$. As a result, $${\rm LHS}_n - {\rm RHS}_{n} = {\rm LHS}_{n-1} - {\rm RHS}_{n-1}$$ and the expression ${\rm LHS}_n - {\rm RHS}_{n}$ is independent of $n$. Since this difference vanishes at $n = 0$, we can conclude ${\rm LHS}_n = {\rm RHS}_n$ for all $n$ and hence establishes the identity at hand.