Show that $\sum_{i=1}^ni^3=\left(\sum_{i=1}^ni\right)^2$ without evaluating them.

Question

Show that $\displaystyle\sum_{i=1}^ni^3=\left(\sum_{i=1}^ni\right)^2$ without evaluating them.

Of course, they are $\displaystyle n^2(n+1)^2/4$ and $n(n+1)/2$, but is there anymore direct way to show that $\sum i^3=(\sum i)^2$ ?

Thank you.