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 $n$th 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 $n$th 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.
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.