I was told there was a proof to this, but it's beyond me. Maybe you guys can either explain why to me or point me in the right direction. I'll apologize in advance for the formatting. It was the best I could do to show the tables.

So, let x^1 = y

x | y

1 | 1
..............1
2 | 2
..............1
3 | 3

By subtracting the y terms, I get one number that stays the same. The number in this case is 1, and that is also the highest exponent from the equation factorial (1!).

Let x^2 = y

x | y

1 | 1
...........3

2 | 4.............2

............5

3 | 9............2

..............7

4 | 16


2! = 2. Just like in the last case. Also, notice I had to go out another level in my differences.

x^3 = y

x | y

1 | 1

...............7

2 | 8..............12

..............19............6

3 | 27............18

...............37...........6

4 | 64............24

................61...........6

5 | 125 ........ 30

.................91

6 | 216


3! = 6, of course.

And, if you work them out, x^4 will equal 24, x^5 equals 120, and so on. I haven't been able to prove this though, because I can't find a universal algebraic statement for it. The parentheses go bonkers on me; the equation for x^3 is not the same as x^4.

So, I turn to you guys. I've looked on google scholar for things like "n! + polynominal" but I can't seem to find anything. Help?

To get better search results, try including the words "finite difference".

Now as for how to prove what you're talking about. For starters, some notation: if S is a sequence, then we'll denote the sequence of differences of S by dS, and the sequence obtained after applying this operator n times by d^n(S).

Try using induction. Clearly, d{x^1}=1!. If we assume that d^n{x^n}=n!, what can we say about d^(n+1){x^(n+1)}? Try writing as d^n((x+1)^(n+1)-x^(n+1)), and see where that leads you. Properties you might want to use (prove them first of course):

d(S+P)=dS+dP, for sequences S and P,

d(aS)=a*d(S), for a sequence S and a real number a,

d^n(x^k)=0 if n>k.

If you happen to know calculus, you should notice that this finite difference operator is a discrete analogue of the derivative.

Thanks for replying.

I've toyed with proving it through mathematical induction, but the problem is that, in order to prove it, I wanted some tangible algebraic equation, but, because of the properties of the "finite difference," the damned equation kept on changing it's parentheses (subtracting the difference of the difference of the difference...), and I had no idea how to set it up for the nth term. However, I will try displaying as a function like you suggested, rather than writing it all out. Thanks for your help.

I do happen to know a little calculus, as in, rules for taking derivatives, and I think I might see what you're talking about.