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 $ $n \in \mathbb{N}$

Here are some interesting observations, which are too long to be included as a comment.  
A useful reference can be found here.
Denoting $\displaystyle\sum_{r=1}^n r^m=\sigma_m$, we note that
$$\begin{align}\sigma_3&=\sigma_1^2\tag{1}\\
\frac{\sigma_5}{\sigma_3}&=\frac {4\sigma_1-1}{3}\tag{2}\\
\frac {\sigma_7}{\sigma_3}&=\frac {6\sigma_1^2-4\sigma_1+1}3\tag{3}\end{align}$$
Adding $(2),(3)$ and using $(1)$ gives 
$$\begin{align}
\frac {\sigma_5+\sigma_7}{\sigma_3}&=2\sigma_1^2=2\sigma_3\\
\sigma_5+\sigma_7&=2\sigma_3^2=2\sigma_1^4\end{align}$$
