I figure why not, to start off the new year. I observed years ago that the sums of the cubes of the first N digits is the sum of those same N digits squared. You can include 0 if you like.

1^3 = 1^2 = 1
1^3 + 2^3 = (1 + 2)^2 = 9
1^3 + 2^3 + 3^3 = (1 + 2 + 3)^2 = 36
1^3 + 2^3 + 3^3 + 4^3 = (1 + 2 + 3 + 4)^2 = 100

and so on forever. However, I have never seen anyone do a good job proving the formula. Can Jemediah or anyone else do it as an exercise?