^{n}where x,n ∈ N can be represented as an nth-order sum where the innermost factor is n!.

First, consider x

^{1}: the difference between x

^{1}and (x+1)

^{1}is 1. So we have a first-order sum where the innermost factor is 1! (= 1.)

x^1 = sum(i = 0..x, 1)

Now let's look at x

^{2}. Let's write the sequence of x

^{2}from -3 to +3 and mark under each two elements the difference between them:

9 4 1 0 1 4 9

-5 -3 -1 1 3 5

2 2 2 2 2

From this we see that we have a second-order sum where the innermost factor is 2! (=2) and the outer factor is -1 (going diagonally left from zero.)

x^2 = -1*sum(i = 0..x, 2*i*sum(j = 0..i, 1))

We can now give a recursive function for the sums that takes a list of factors as its argument (this one only works for positive integers though):

sumMap lst f = sum (map f lst)

sums [] i = 0

sums [x] i = x * i

sums (x:xs) i = x*i + sumMap [0..i] (sums xs)

Then, let's look at the left side of x

^{6}:

46656 15625 4096 729 64 1 0 1

-31031 -11529 -3367 -665 -63 -1 1

19502 8162 2702 602 62 2

-11340 -5460 -2100 -540 -60

5880 3360 1560 480

-2520 -1800 -1080

720 720

From which we get the factors [-1, 62, -540, 1560, -1800, 720] and

x^6 = -1*sum(a=0..x, 62*sum(b=0..a, -540*sum(c=0..b, 1560*sum(d=0..c, -1800*sum(e=0..d, 720*sum(f=0..e, 1))))))

or

pow6 = sums [-1, 62, -540, 1560, -1800, 720]

In the general case, the first factor is -1

^{(n-1)}, the second factor is -2

^{n}+ 2 * -1

^{n}, the second to last factor is -(n-1)n! * 2

^{-1}and the last factor is n!. I don't know about the rest of the factors.

### Fun with cubes

Here's the difference grid for the third power:

-27 -8 -1 0 1 8 27

19 7 1 1 7 19

-12 -6 0 6 12

6 6 6 6

Which gives the factors [1, -6, 6].

One result of the sum representation is that each x cubed minus x is divisible by 6:

`6 | x`^{3} - x

. Or put another way, `x`^{3} = x + 6k, k ∈ Z

. And the sum between two cubes minus the sum of the bases is also divisible by six: `6 | x`^{3} + y^{3} - (x + y)

.We also see that the difference between any two integer cubes grows as a second order function:

`n`^{3} - (n-1)^{3} = 6(n/2)(n+1)+1 = 3n^{2}+3n+1

.

cubeDiff n = 3*(n^2 + n) + 1

pow3 n = sumMap [0..n-1] cubeDiff

-- pow3 5

-- 125

There's another amusing difference grid in

`(x`^{3} - x) / 6

:

0 1 4 10 20 35 56 84

1 3 6 10 15 21 28

2 3 4 5 6 7

1 1 1 1 1