Find variance of sample mean in simple random sampling without replacement.

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Hi I'm trying to variance of sample mean in simple random sampling without replacement(finite population correction factor).

And my reasoning is this; Let $I_i$ be a indicate variable of a event that ith element is in a sample of n. So $E[I_i]=n/N$ and $Cov(I_i,I_j)=n/N-n^2/N^2$ if $i=j$ and $Cov(I_i,I_j)=n(n-1)/N(N-1)-n^2/N^2$. Let $\overline X=\frac1n \sum_1^nx_iI_i $. Then $E[\overline X]=\sum_1^nx_i/N=\bar x$. And $Var(\overline X)=Var(\frac1n \sum_1^nx_iI_i)=\frac{1}{n^2}Var(\sum_1^nx_iI_i)=Cov(\sum_1^nx_iI_i,\sum_1^nx_jI_j)=\sum\sum_{i=j}Var(x_iI_i)+2\sum\sum_{i<j}Cov(x_iI_i,x_jI_j)$.

But I can't go futher than this. How could i coclude $Var(\overline X)=\frac{N-n}{N-1}Var(X)$?

Thanks in advance.