<s>[INST] Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.,
Is
$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4$$ true $\forall $ $n \in \mathbb{N}$
 [/INST] 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.
</s>