Prove that successive differences of powers of integers a^n end up with n!?

Question

Consider series of squares 1, 4, 9, 16, 25... The differences are 3, 5, 7, 9... which leads to differences of 2, 2, 2, 2... = 2! The series of cubes 1, 8, 27, 64, 125... similiarly ends up with 6, 6, 6, 6 ... = 3! Etc. It always ends up with n!, n!, n!,.... where n is the power of the original series.

I'm going to leave this question open to see if anyone can offer a non-calculus proof.

Thanks, Zanti3, for your non-calculus answer. Now I have the really hard job of picking one of you as the best answer. Maybe I should let voters decide?

In fact, let's say that if a chart of successive differences is displayed, so that the first value of the top row is a, the 2nd row b, 3rd c, etc, then the terms of the series at top is the sum a + bn + c(1/2!)(n-1)n + d(1/3!)(n-2)(n-1)n + e(1/4!)(n-3)(n-2)(n-1)n +... etc. In other words, a Pascal's "field" instead of a triangle.

Answer 1

OK, I think I can do this with pre-calculus math:

Given a polynomial function P(x) of degree k, then taking the difference of consecutive terms of the sequence P(1), P(2), P(3), P(4), ... produces a sequence for a polynomial function of degree k-1. Repeat this k+1 times, producing k+1 rows, and you will get to a row of constants. Conversely, if you have a function that reduces to constants after creating k+1 rows, then your function is a polynomial of degree k.

I think the easiest way to see this is to work it backwards. Let's say the row of constants is a, a, a, a, .... Then, if b is the first term in the row above the row of constants, then the terms of this row are b, b+a, b+2a, b+3a, ..., and the nth term is a(n-1) + b. This is a polynomial of degree 1.

Similarly, if c is the first term in the third row, then the terms of this row are c, c+b, c+2b+a, c+3b+3a, c+4b+6a, ..., and the nth term is a(n-1)(n-2)/2 + b(n-1) + c, a polynomial of degree 2. Continuing in this way, it becomes apparent that, for the (k+1)th row, the nth term is:

a(n-1)(n-2)...(n-k)/k! + b(n-1)(n-2)...(n-k+1)/(k-1)! + c(n-1)(n-2)...(n-k+2)/(k-2)! + ... (There will be k+1 terms in all.)

This gives us a neat way to find the polynomial function for a sequence if we don't know it: take differences until we get to the row of constants, then plug the first numbers in each row into the above equation. For your question, however, we know the polynomial function is n^k. So, the row of constants is the (k+1)th row,and we know that:

n^k = a(n-1)(n-2)...(n-k)/k! + b(n-1)(n-2)...(n-k+1)/(k-1)! + c(n-1)(n-2)...(n-k+2)/(k-2)! + ...

Since we know this formula is valid, somehow all the terms on the right hand side must cancel out, leaving only n^k. Also - the key point for answering this question - the only term that contributes to the n^k part of the polynomial is the first term, i.e. the term with the a coefficient. Therefore, for the right hand side to simplify to n^k, a = k! is necessary.

Answer 2

since delta x is synonymous to dx... the relationship is affected by the derivative..

consider x^2 ... first derivative = 2x
second derivative = 2

x^3 ...
first derivative = 3x^2
second derivative = 6x
third derivative = 6

consequently
for x^n ... the nth derivative = n!

when you continue with the finite differences.. you will eventually arrive at some constant value (whenever you have a polynomial)... this constant value is the value of the nth order derivative for some n of that polynomial.

in the other direction...
getting the finite differences for certain series of numbers, arranged in a fixed interval... you will obtain the equation of the polynomial fitting the set of numbers.