Sum of the cubes of the first $n$ natural numbers I know that the sum of the cubes of the first $n$ natural numbers is  $\{\frac {n(n+1)}2\}^2$ but I am looking for a method to derive this. 
If there is a method please try to make as simple as possible and possibly without the use of mathematical induction or calculus. 
I apologize if I sound ungrateful but I am looking for a way to derive it without using anything to do with Binomial Theorem, Calculus, Mathematical Induction or the fancy notation that almost all answers contain. 
If there is no such method then can someone explain what the notation on top of and on the bottom of the Sigma symbol? Like this one $\sum_{i=1}^n$  I do not understand what this means. 
I do know how the sum of the first $n$ natural numbers using the Gaussian Method as a derivation, however, I am not sure even about the sum of the first $n^2$ natural numbers. 
 A: $$(n+1)^4-1^4=\sum_{i=1}^{n} (i+1)^4-i^4=4\sum_{i=1}^{n}i^3+6\sum_{i=1}^{n}i^2+4\sum_{i=1}^{n}i+\sum_{i=1}^{n}1$$
Assuming you know that: $\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$, $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$, $\sum_{i=1}^{n}1=n$ we can plug this in and solve:
$$4\sum_{i=1}^{n}i^3=(n+1)^4-1-6\left (\frac{n(n+1)(2n+1)}{6}\right )-4\left (\frac{n(n+1)}{2}\right )-n=n^2(n+1)^2$$
And thus 
$$\sum_{i=1}^{n}i^3=\frac{n^2(n+1)^2}{4}=\left ( \frac{n(n+1)}{2}\right)^2$$
Edit: 
a nice picture from Wikipedia:
Notice the pattern between even and odd numbers.
A: If you already know the sum $S_1$ of the $n$ natural numbers, and the sum $S_2$ of their squares, you can easily derive it:
Start from the binomial formula $\,(k+1)^4=k^4+4k^3+6k^2+4k+1$, which we rewrite as
$$(k+1)^4-k^4=4k^3+6k^2+4k+1$$
Summing it for $k=1$ to $n$, we obtain
$$(n+1)^4-1^4=4S_3+6S_2+4S_1+n,$$
whence the formula
$$S_3=\frac14\bigl((n+1)^4-(n+1)-6S_2-4S_1\bigr).$$
A: You can derive it from the fact that the sum of the first $k$ odd numbers (up to $2k-1$) is $k^2$. That is, $k^2 = \sum_{i=1}^k (2i - 1)$.
We can write $k^3$ = $k\cdot k^2 = \sum_{i=1}^k k^2 = \sum_{i=1}^k k^2 + (\sum_{i=1}^k (2i-1) - k^2) = \sum_{i=1}^k (k^2 - k - 1 +2i) $
This is the sum of all odd numbers from $k^2 - k +1$ through $k^2 + k - 1$.
The next odd number is $k^2 + k + 1 = (k+1)^2 - (k+1) + 1$, i.e., the first odd number contributed by the $(k+1)^3$ term.
Thus the sum of first $n$ cubes is the sum of the odd numbers through $n^2 + n - 1 = 2\frac{n(n+1)}{2} - 1$, which is $\left[\frac{n(n+1)}{2}\right]^2$.
A: I do not think that this is correct.
$\frac{n(n+1)}{2}=\sum_{i=1}^n i$.  
For a dervation on that you can look at it like this:
Let $S=\sum_{i=1}^n i$. Then $2S=\sum_{i=1}^n 2i=n+\sum_{i=1}^n (i+n-i)=n(n+1)$, therefore $S=\frac{n(n+1)}{2}$
