summation - Why is $(1+2+3+4+...+n)^2$ equal to $1^3+2^3+3^3...+n^3$?

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?