Prove that If $p$ is a prime then $p$ divides $(a+b)^p-a^p-b^p$ I have to prove 

If $p$ is a prime then $p$ divides $(a+b)^p-a^p-b^p$, with $a$, $b \in \mathbb N$.

My approach:
$$(a+b)^p=\sum \binom{p}{r}a^{n-r}b^r$$
where $r$ goes from $1$ to $p$.
Then $a^p$ and $b^p$ would be cancelled and remaining term would be $\displaystyle \binom{p}{r}$ which is a multiple of $p$ if $p$ is a prime.
Otherwise if $p$ is not a prime then it would be cancelled from the factor of denominator
since it is a prime it would not exist in the factorials of no less than $p$ 
Hence we have proved.
Am I right?
Can anyone suggest me any alternate way?
 A: You have the correct ideas, however the presentation may not be understandable by someone who doesn't already know how this problem "works".
First, as a matter of simple error, the index $r$ in your summation formula goes from 0 to $p$.
What you correctly stated and which needs to be proven is that ${p \choose r}$ is divisible by $p$ if $p$ is a prime and $1 \le r \le p-1$.
You stated: 

Otherwise if p is not a prime then it would be cancelled from the factor of denominator.

Thinking about what is different about ${p \choose r}$ when $p$ is a prime vs. when it is not might be important when trying to solve problem, but stating it here serves no purpose, because we are only dealing with a $p$ that is prime. It is also not correct, as ${p \choose 1} = p$ is a value divisible by $p$ for all $p$. Similiarly, ${9 \choose 2} = \frac{9*8}{2}= 36$ is divisble by 9, etc.
The correct argument that you gave is 

since it is a prime it would not exist in the factorials of no less than p 

More formally, since ${p \choose r} = \frac{p!}{r!(p-r)!}$ the enumerator is divisible by $p$, but since $r < p$ and $p-r < p$ (remember $1 \le r \le p-1$), none of the factorials in the denominator is divisible by $p$, as $p$ is prime.
A: Okay, So you want confirmation/verification of your proof.
I must say you are right. But notice one thing that you are using a long method, there is no need to take the case when $p$ is not prime. The question ask that $p|(a+b)^p-a^p-b^p$ if P is prime. So, the moment when you prove that every prime p divides $(a+b)^p-a^p-b^p$ you are done.
In the expansion $(a+b)^p=\sum \binom{p}{r}a^{n-r}b^r$ the terms $a^p$ and  $b^p$ will be cancelled out and the remaining terms will contain $\displaystyle \binom{p}{r}$, where the relation $1 \le r \le p-1$ holds. Since $p$ is prime, no value from the set {$1,2,3,......p-1$} will divide $p$, Or $\displaystyle \binom{p}{r}$ will still contain $p$ after arithmatical operations. so $p$ will divide every term of the form $\displaystyle \binom{p}{r}$. 
Hence $p$ will divide every term of the expansion $(a+b)^p=\sum \binom{p}{r}a^{n-r}b^r$ (except $a^p$  and $b^p$  which are already cancelled out). I think we are done.
