here's an alternative geometric proof
imagine you have n cubes. the ith cube is made up of i^3 little unit cubelets.
we want to show that we can rearrange these cubelets into a square with side length 1 2 ... n.
the base case is trivial. assume for induction that we have already shown this for n. then, consider taking a cube of side length n 1 and slicing it up into n 1 new squares, each of side length n 1, now we want to arrange these n 1 side length squares around the original square to produce a larger square. how can we do that?
the original square has side length n(n 1)/2. in other words, you can fit n/2 copies of n 1 side length squares along each side of the original square. do this on the upper edge and the right edge, say. and there's a gap in the upper right hand corner that can exactly be filled with one more n 1 side length square. in other words we have just placed exactly n 1 copies of n 1 side length squares, creating a new perfect square that has a side length n 1 greater than the original one.