What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression using sums $\sum$ and double sums $\sum$$\sum$?

Here is my favorite method which works for any polynomial summand and you only need to remember two basic facts, one from calculus one and one about polynomials. First, since summations are analogous to integration, we have that
$$\int x^k \approx x^{k+1} \Rightarrow \sum x^k \approx x^{k+1}.$$
For your problem, let us define
$$f(n)=\sum_{i=1}^n i^4$$
and since the summand is a polynomial of degree four, the sum $f(n)$ must be a polynomial of degree five. If you don't believe me then just compute at least the first six values of $f(n)$ and compute the sixth differences and all of the terms will be zero (analogous to the sixth derivative of a fifth degree polynomial being zero).
Then using the (second) fact that a polynomial of degree five can be uniquely determined by six points, use points
$$(1,f(1)),(2,f(2)),(3,f(3)),(4,f(4)),(5,f(5)),(6,f(6))$$
and compute the unique interpolating polynomial and you get
$$f(n)=\frac{6n^5+15n^4+10n^3-n}{30}.$$
