How to prove that nth differences of a sequence of nth powers would be a sequence of n!

Given an infinite sequence of numbers, first differences denote a sequence of numbers that are pairwise differences, second differences denote a new sequence of pairwise differences of this sequence, and so on.

1  2  4  7 11 16 22 29
1  2  3  4  5  6  7    -- first differences
1  1  1  1  1  1      -- second differences
0  0  0  0  0        -- third differences


Sequences of nth powers of numbers exhibit an interesting property. Take a sequence of squares: (1,4,9,16,25,...).

• First differences: (3,5,7,9,11...)
• Second differences: (2,2,2,2,2...)

Take a sequence of cubes: (1,8,27,64,125...)

• First differences: (7,19,37,61,91...)
• Second differences: (12,18,24,30,36...)
• Third differences: (6,6,6,6,6...)

Take a sequence of powers of 4: (1,16,81,256,625,...)

• First differences: (15,65,175,369,671,...)
• Second differences: (50,110,194,302,434,...)
• Third differences: (60,84,108,132,156,...)
• Fourth differences: (24,24,24,24,24,...)

1, 2, 6, 24... that's clearly a factorial! So a sequence of powers of n will have a sequence of repeating n! as its nth differences. How do I prove this relation holds? Bonus points if you give just enough theoretical information for a non-mathematician, but leave the proof as an exercise.

Note: It actually doesn't matter if our sequence includes negative numbers too, sequence (...,-2,-1,0,1,2,...) still has (...,1,1,1,1,1,...) as its first differences.

• True, of course. See if you can prove it for exponents $1,2,3.$ – Will Jagy Dec 1 '14 at 4:13
• You should try to build a theory of "discrete differential and integral calculi" where real valued functions are replaced with sequences, the derivative is replaced with the difference operator, and the integral is replace with the summation operator. You even get an analog of the fundamental theorem of calculus. – Baby Dragon Dec 1 '14 at 4:15

The difference sequence of any polynomial $Cx^n + \cdots$ (where here and elsewhere, $\cdots$ hides terms of lower degree) is of the form $Cnx^{n-1} + \cdots$, since $(x+1)^n - x^n = nx^{n-1} + \cdots$ (just open the binomial). If we start with $x^n + \cdots$ then we get $nx^{n-1} + \cdots$, then $n(n-1)x^{n-2} + \cdots$, then $n(n-1)(n-2)x^{n-3} + \cdots$, and eventually, after $n$ times, $n(n-1)(n-2)\cdots 1 x^0 = n!$. You can prove this by induction.