How to prove this rule?

[ChretienDeTroyes]

Hey,
I wanted to prove the rule:
(x+1)^p - (x^p) -1| p
x > 0, x ∈ ℕ
p = prime number

So far I have no experiences with proving divisibility, which is why I need help here.

 

[Zermelo]

Hi there, shouldn't it be the other way around? $$p \lvert (x+1)^p - x^p - 1$$?
What does divisibility mean? By definition, $a \lvert b \Longleftrightarrow ( \exists k \in Z ) b = a \cdot k$, so basically, if I'm right about it being the other way around, you need to prove that p is a factor of $(x+1)^p - x^p - 1$, aka $(x+1)^p - x^p - 1$ is p times some integer k (the expression for k can be really complicated, but it's only important that you know that k is an integer, no matter how complicated the expression is. Note that multiplication and addition are closed operations in Z - the set of all integers, aka, when you multiply and add some integers, the result is an integer!). Start with $(x+1)^p - x^p - 1$, try to calculate it, use the binomial formula, it's not that hard believe me! When you're done simplifying, you should be able to factor out p out of the resulting expression

EDIT:
Check out "Fermat's little theorem!" I think it could be useful here!.
Also, if those 2 ways don't work, you could try with mathematical induction on x! Prove that it holds for x = 1, assume that it works for x = n, and try to prove it works for x = n+1.
Those are some standard ways of proving divisibility.

 

[ChretienDeTroyes]

Thanks for your help.