<s>[INST] Proof by induction (Combinatorics) This is my first question on here so please bear with me, thank you.

Prove that for all positive integers $n$
  $$2(1+2+...+n)^4 = (1^5 + 2^5 +...+ n^5) + (1^7 + 2^7 +...+ n^7)$$

After establishing the base case, I proceeded to the induction:
$2(1+2+...+(n+1))^4 = (1^5 + 2^5 +...+ (n+1)^5) + (1^7 + 2^7 +...+ (n+1)^7)$
$2(1+2+...+(n+1))^4 = (1^5 + 2^5 +...+ n^5)+(n+1)^5 + (1^7 + 2^7 +...+n^7)+(n+1)^7$
$2(1+2+...+(n+1))^4 = 2(1+2+...+n)^4 + (n+1)^5 + (n+1)^7$
$2(1+2+...+(n+1))^4 - 2(1+2+...+n)^4 = (n+1)^5 + (n+1)^7$
$2[(1+2+...+(n+1))^4 - (1+2+...+n)^4] = (n+1)^5 + (n+1)^7$
And from here I've hit a wall and can't figure out how to continue. I tried substituting $(1+...+n) = [n(n+1)]/2$ and $(n+1)=[2(1+...+n)]/n$ but to no avail.
 [/INST] Note that
$$1+2+\cdots+n =\frac{n(n+1)}{2},$$ and $$1+2+\cdots+n+1 =\frac{(n+1)(n+2)}{2}.$$
Therefore, the left-hand side of the last equation is
$$\frac{(n+1)^4}{8}\left((n+2)^4 -n^4\right)=\frac{(n+1)^4}{8}\left((n+1+1)^4 -(n+1-1)^4\right).$$
But $$(a+1)^4-(a-1)^4 = ((a+1)^2)^2-((a-1)^2)^2\\=((a+1)^2-(a-1)^2)((a+1)^2+(a-1)^2) \\=(a+1+a-1)(a+1-a+1)(a^2+1+2a+a^2+1-2a)\\=8a(a^2+1).$$
Using the above relationship for $a=n+1$, $$\frac{(n+1)^4}{8}\left((n+2)^4 -n^4\right)=(n+1)^4(n+1)((n+1)^2+1) = (n+1)^5+(n+1)^7.$$
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