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I noticed that the sum of the first $n$ cubes is equal to the square of sum of the first $n$ natural numbers: $$ \sum\limits_{i=1}^n i^3=\frac{n^2(n+1)^2}{4}=\left(\frac{n(n+1)}{2}\right)^2=\left(\sum\limits_{i=1}^n i\right)^2 $$

Is there a clever way to explain (not prove) this identity?

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