The formula for the nth triangular number is T[sub]n[/sub] = ½n(n+1), from this it is easy to show that T[sub]n[/sub][sup]2[/sup] - T[sub]n-1[/sub][sup]2[/sup] = n[sup]3[/sup]

Here’s the simplest proof I know, though others will surely feel differently:

Consider 1[sup]p[/sup] + … + n[sup]p[/sup] for integer n. Note that this is always given by a (p+1)-th degree polynomial in n with leading coefficient 1/(p + 1), sending 0 to 0 and 1 to 1; furthermore, this polynomial will be symmetric around -0.5 for odd p and antisymmetric around -0.5 for nonzero even p.

Thus, T[sub]3/sub and (T[sub]2/sub)[sup]2[/sup] are both given by 4th degree polynomials with leading coefficient 1/4, sending 0 to 0 and 1 to 1, symmetric around -0.5. This leaves no degrees of freedom for them to differ, and thus they are the same.

Whoops, I made a mistake in editing: I meant to say “Let T[sub]p/sub be 1[sup]p[/sup] + … + n[sup]p[/sup] for integer n”, so I could refer to that name again later.