The sum of the cubes of the first n natural numbers

We shall prove that

$$\sum_{k=1}^{k=n}k^3=1^3+2^3+3^3+...+n^3=\frac{n^2(n+1)^2}{4}$$

First proof is by induction.
For n=1 we have

$$1^3=\frac{1^22^2}{4}=1$$

We shall prove that

$$\sum_{k=1}^{k=n+1}=1^3+2^3+3^3+...+n^3+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$$

or

$$\sum_{k=1}^{k=n}k^3+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$$

But from induction supposition we have to prove

$$\frac{n^2(n+1)^2}{4}+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$$

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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