Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site 
states: 

Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
  “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.”
In other words, according to Example $1$: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}.$$

Should:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}$$
not be:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^3 +(n + 1)^3}{2^3}$$
as everything in the left-hand side is cubed?

Argument:
Every $k^3$ is the sum of $k$ consecutive odd numbers.
E.g.: $8 = 3+5$, $\quad 27 = 7+9+11$, $\quad 64 = 13+15+17+19$.
The sum of $n$ consecutive $k^3$ numbers, starting from $k = 1$, is the sum of
$n(n+1)/2$ consecutive odd numbers. 
But the sum of a number of consecutive odd numbers is the square of 
that number.
Therefore, the sum of $n$ consecutive $k^3$ starting with $1$ is $[n(n+1)/2]^2$.
Elaboration:
It is well known that $k^2$ is the sum of $k$ consecutive odd numbers starting from $1$.  But $k^3$, $k^4$, and $k^p$ generally, are also the sum of $k$ consecutive odd numbers.  This is because $k$ numbers whose arithmetic mean is $k$ sum to $k^2$; $k$ numbers whose mean is $k^2$ sum to $k^3$, et cetera.  What is unique about the sum of cubes is that it is a sum of consecutive odd numbers with no gaps and no repetitions.  E.g., the sum of the first five cubes is:
$$\underbrace{1^3}_{1} +\underbrace{2^3}_{3+5} +\underbrace{3^3}_{7+9+11} +\underbrace{4^3}_{13+15+17+19} +\underbrace{5^3}_{21+23+25+27+29}.$$
Therefore the sum is a square.  But what square?  In summing the first five cubes we have summed the first $1+2+3+4+5 = 15$ odd numbers, giving us $15^2$.  And $15$ is the fifth triangle number.  Hence generally, the sum of the first $n$ cubes is the square of the $n$th triangle number, or $[n(n+1)/2]^2$, i.e. the square of the sum of the first $n$ integers.
With sums of squares, there is repetition because all the sums of odd numbers start from $1$: 
$$1^2 +2^2 +3^2 +4^2 = 1 +(1+3) +(1+3+5) +(1+3+5+7).$$
With fourth and higher powers there is no repetition, but there are gaps between the sets of odd numbers being summed:
$$1^4 +2^4 +3^4 +4^4 = 1 +(7+9) +(25+27+29) +(61+63+65+67).$$
Only sums of cubes are sums of consecutive odd numbers starting from $1$,
and therefore have this special connection with squares.
