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.

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