Show that the ring $R$ is commutative with unity with $x^n=x$ for all $x\in R$ has finite characteristic. Suppose that $R$ commutative ring with unity and for all $x\in R$, $x^n=x$  and $n$ is fixed positive integer. Show that $R$ has finite characteristic. 
How to prove that for unknown integer $n$. Also what is the possible numbers for characteristic . Is it the divisors of $n$.
I tried to start like this way : 
$2x=(2x)^n=2^nx^n \Rightarrow (2^n-2)x=0 $. So this prove $R$ has finite characteristic, right ? but that will give me the possible values for characteristic for  $R$ is the divisors of $2^n-2$. I am so confused . I appreciate any hint. Thanks
I know in this type of ring every prime ideal is maximal, reduced, and Von Neumman ring. 

Hint:
$$(1+1)^n=1+1$$
Therefore,
$$\sum_{j=1}^{n-1}\binom nj=0$$
