Proof that sum of first $n$ cubes is always a perfect square I know that
$$1^3+2^3+3^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2$$
What I would like to know is whether there is a simple proof (that obviously does not use the above info) as to why the sum of the first $n$ cubes always a perfect square.

Let's prove this quickly by induction.
If needed I will edit this answer to provide further explanation.
To prove: $$\sum_{i=1}^n i^3=\left(\frac{n(n+1)}{2}\right)^2$$
Initial case $n=1$: $$\sum_{i=1}^1 i^3=1^3=\left(\frac{2}{2}\right)^2 = \left(\frac{1(1+1)}{2}\right)^2$$
Given for induction: $$\sum_{i=1}^n i^3=\left(\frac{n(n+1)}{2}\right)^2$$
To show in induction step: $$\sum_{i=1}^{n+1} i^3=\left(\frac{(n+1)(n+2)}{2}\right)^2$$
Induction step:
$$
\sum_{i=1}^{n+1} i^3=(n+1)^3+\sum_{i=1}^{n} i^3\stackrel{!}{=}
(n+1)^3+\left(\frac{n(n+1)}{2}\right)^2
= \left(\frac{4(n+1)^3+(n(n+1))^2}{2^2}\right)
= \left(\frac{4(n+1)(n+1)^2+n^2(n+1)^2}{2^2}\right)
= \left(\frac{(n+1)^2(4(n+1)+n^2)}{2^2}\right)
= \left(\frac{(n+1)^2(n^2+2\cdot2n+2^2)}{2^2}\right)
= \left(\frac{(n+1)(n+2)}{2}\right)^2
$$
In $(!)$ we used the equation holding for $n$.
If you are not familiar with the principle of induction, take a look at this.
