Hi there, shouldn't it be the other way around? [math]p \lvert (x+1)^p - x^p - 1[/math]?
What does divisibility mean? By definition, [imath]a \lvert b \Longleftrightarrow ( \exists k \in Z ) b = a \cdot k[/imath], so basically, if I'm right about it being the other way around, you need to prove that p is a factor of [imath] (x+1)^p - x^p - 1[/imath], aka [imath] (x+1)^p - x^p - 1[/imath] 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 [imath] (x+1)^p - x^p - 1[/imath], 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.
