Remember the definition:

andWe can continue this and notice a pattern:

and

So in general:

What we want to prove is that:

To do so, we will prove that

and then look at the case where and we already proved that whenAgain we use induction:

We must assume that

Then

Expanding:

We see pairs of

and

Now we rewrite our previous expansion as:

Every k has been replaced with k+1, thus proving that

Now for the best part! We shall look at the case where k=n:

Substituting n for k in our sum:

Now, x is arbitrary; it does not matter where in a sequence of perfect powers of n we begin taking differences, so we will let x=0:

That is my proof Let me know if you guys see any errors (maybe I am completely wrong). Also, I apologize if it is a messy proof either due to my lack of LaTeX skills or lack of organization. Off to bed now! Bye.

]]>Consider the difference pyramid discussed earlier in the thread. Each row is the difference between consecutive terms of the previous row. We will define a function to describe the differences:

. Here represents the number of times we have taken the difference, is a number in a row, and is the exponent. By this, we can define the first row (the 0th difference) with . Thus, by our definitions, . This should make sense because is the first difference between two consecutive powers of n.We can expand this:

Now we can apply the same process for further differences:

This is where we use induction: we want prove that the nth difference is equal to n!

To do this we need to assume two things:

First, we assume that the nth difference is, in fact, equal to n!, and we need to assume that after we reach a row of n!, the next difference is 0.

Assume:

Now we prove this for

:This was just the first part of my proof, showing that the nth difference of perfect powers of n is equal to n!. I will use this to prove that n! can be expressed as an alternating sum. I have to go for now, so I will post that part later. Bye for now! And let me know if I made any errors or did anything improperly. I didn't catch any mistakes, but I'm sure you guys can if I made any.

]]>Why not make another Introduction thread?

]]>No problem, I knew who you were. Welcome to the forum.

]]>To do it, I had to create a function of two variables and prove that the nth difference of perfect powers of n is n!. Then I used that result to show that the nth difference can be expressed by the alternating sum, thus proving that n! is equal to the alternating sum. If anybody would like to see, I could practice my LaTeX skills and write it up on here some time

Also, I have been reading the forum for quite some time now. I really enjoy it, so I will probably be spending some time here. I guess this is sort of an introduction, so hello all!

]]>It is a nice result anyway is the correct attitude.

Do you use Maple alot? Do you like it?

I am going to take a break, see you later.

]]>Is your formula more efficient for a computer then the normal one for say 100! ?

]]>Could it have any practical uses?

]]>I would say yes but so what. If you discovered it on your own then it is unique and new for you.

]]>It does work on all the examples I gave it. Here is what it looks like in latex.

]]>Substuting 4 in for n in sum((-1)^p*nCp*(n-p)^n, p=0..n) gives sum((-1)^p*4Cp*(4-p)^4, p=0..4).

Taking the sum gives (-1)^0*4C0*(4-0)^4 + (-1)^1*4C1*(4-1)^4 + (-1)^2*4C2*(4-2)^4 + (-1)^3*4C3*(4-3)^4 + (-1)^4*4C4*(4-4)^4

= 1*1*256 + (-1)*4*81 + 1*6*16 + (-1)*4*1 + 1*1*0

= 256 - 324 +96 -4 +0

=24

]]>Could you show a small example with numbers?

]]>