<s>[INST] 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?
 [/INST] $$1^3 +2^3 +\ldots +n^3 =\left(\frac{n(n+1)}{2}\right)^2.$$
We will prove by induction on $n$, that
$$\sum_{k=1}^n k^3 = 1^3 +2^3 +\ldots +n^3 =\frac{n^2 (n+1)^2}{4}.$$
For $n=1$, we have $1^3 = \dfrac{1^2 2^2}{4} = 1$.
We shall prove that
$$\sum_{k=1}^{n+1} k^3 = 1^3 +2^3 +\ldots +n^3 +(n+1)^3 =\frac{(n+1)^2 (n+2)^2}{4},$$
by assuming that
$$\sum_{k=1}^n k^3 = 1^3 +2^3 +\ldots +n^3 =\frac{n^2 (n+1)^2}{4}.$$
From the induction supposition, we have to prove that
$$\frac{n^2 (n+1)^2}{4} +(n+1)^3 = \frac{(n+1)^2 (n+2)^2}{4},$$
or
$$(n+1)^3 = \frac{(n+1)^2 (n+2)^2}{4} -\frac{n^2 (n+1)^2}{4} = \frac{(n+1)^2}{4} ((n+2)^2 -n^2) = (n+1)^3.$$
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