We will prove this by induction.
When there is nothing to prove.
So say that, .
We will prove it for .
Here is a useful result: if *.
With this result we get, .
However, ... 
Here is another result: if then divides .**
Thus every term in sum of  for is divisble by notice that .
Therefore, each term for is divisible by . For the last term ( ) in sum of  we have . But since . And so we have shown that divides each term for in the sum of . Thus, the sum is equivalent to 0 mod .
*)Argue by induction. If it means for some . Now raise both sides to the power and apply the binomial theorem again. I think the result that appears in ** (just below) will be useful in this proof.
**)Because is an integer, since for we have that divides . Thus, it leaves a factor of . Thus, divides .
We can now prove a stronger result. That the order of is mod . Where .
Okay we have shown . This implies, . But, (use binomial theorem). However, since . The rest follows.