Proving $E(\hat s^2)=V(\bar x)$ for finite population Population is $X_1,X_2,...,X_N$ with large but finite $N$. Sampling indicator $Z_i \in(0,1)~ \forall i$ , such that $\sum^N _{i=1}Z_i=n$ and $Pr(Z_i=1)=\frac {n}{N}$. 
Sample mean in this case is $\bar x=\frac{1}{n}\sum^N _{i=1}Z_iX_i$, where $E(\bar x)=\bar X$. 
Variance of the estimator is $V(\bar x)=(1-\frac {n}{N})\frac {S^2}{n}$, where $S^2=\sum^N _{i=1}(X_i-\bar X)^2/(N-1)$ is the population variance. 
Sample variance is $\hat s^2=(1-\frac {n}{N})\frac {s^2}{n}$, where $s^2=\sum^N _{i=1}Z_i(X_i-\bar x)^2/(n-1)$. 
I want to show that $E(\hat s^2)=V(\bar x)$, i.e.,the sample variance is an unbiased estimator. Note that $Z_i$ is the only random variable here, so does $\bar x$, because $\bar x$ consists $Z_i$.
I derived the following but do not know how to proceed foward, could someone help me with this please? Thanks!
My attempt:
\begin{align}
E(\hat s^2) &= \frac {1-\frac {n}{N}}{n(n-1)}E\bigl[ \sum^N_{i=1}Z_i(X_i-\bar x)^2 \bigr] \\
&=\frac {1-\frac {n}{N}}{n(n-1)} E\Bigl[ \sum^N_{i=1} Z_i \cdot \bigl( X_i -\bar X) + (\bar X-\bar x) \bigr)^2 \Bigr] \\
&= \frac {1-\frac {n}{N}}{n(n-1)}E\Bigl[\sum^N_{i=1} Z_i \cdot \bigl( (X_i-\bar X)^2+2(X_i-\bar X)(\bar X-\bar x)+(\bar X-\bar x)^2 \bigr) \Bigr] 
\end{align}
I know that  
$$E[\sum^N_{i=1}Z_i(X_i-\bar X)^2]=\sum^N_{i=1}E(Z_i)(X_i-\bar X)^2=\sum^N_{i=1}Pr(Z_i=1)(X_i-\bar X)^2=\frac{n}{N}\sum^N_{i=1}(X_i-\bar X)^2$$ 
How should I deal with the latter terms of $2(X_i-\bar X)(\bar X-\bar x)+(\bar X-\bar x)^2$? Any comment would be helpful!
 A: The common anchoring $X_i - \bar x = (X_i - \bar X) + (\bar X - \bar x)~$ is very meaningful and can be seen to be useful as per the comments by papasmurfete. Nonetheless, in this case we can arguably have a simpler algebraic derivation without that anchoring. 
Back to the beginning, we expand the square into the usual 3-parts but only bring the expectation inside the summation for the first term.
\begin{align}
\frac{ n(n-1) }{1 - \frac{n}N } E(\hat s^2) &= \color{blue}{ \sum^N_{i=1} E \Bigr[ Z_i X_i^2 \Bigr] } - 2 E \Bigl[ \sum^N_{i=1} Z_i X_i \bar x \Bigr] + E \Bigl[ \sum^N_{i=1} Z_i \bar x^2 \Bigr]  \\
&= \color{blue}{ \sum^N_{i=1} X_i^2 E \Bigr[ Z_i \Bigr] } - 2 E \Bigl[ \bar x \sum^N_{i=1} Z_i X_i \Bigr] +  E \Bigl[ \bar x^2 \sum^N_{i=1} Z_i \Bigr] \\
&= \color{blue}{ \frac{n}N \sum^N_{i=1} X_i^2} - 2 E \Bigl[ \bar x \cdot n \bar x \Bigr] + E \Bigl[ \bar x^2 \cdot n \Bigr] \\
&= \frac{n}N \sum^N_{i=1} X_i^2 - n E \Bigl[ \bar x^2 \Bigr]
\end{align}
Hopefully each step above is self-explanatory. Now, the common factor $n$ cancels on two sides of the equation, and we know the second moment is the variance plus the square of mean.
\begin{align}
\frac{ n-1 }{1 - \frac{n}N } E(\hat s^2) &= \frac1N \sum^N_{i=1} X_i^2 -  \left( V( \bar x^2 ) + E[\bar x]^2 \right) \\
&= \frac1N \sum^N_{i=1} X_i^2 - V( \bar x^2 ) - \bar X^2  \\
&= \frac1N \left( \sum^N_{i=1} X_i^2 - N \bar X^2  \right) - V( \bar x^2 ) \\
&= \frac1N  \sum^N_{i=1} (X_i - \bar X)^2 - V( \bar x^2 ) \\
&= \frac1N \frac{ n (N - 1) }{1 - \frac{n}N } V( \bar x^2 ) - V( \bar x^2 ) 
\end{align}
The coefficient for $V( \bar x^2 )$ on the right hand side is shown to be
$$ \frac{ n (N - 1) }{N - n} - 1 = \frac{ n N - n - (N - n)}{N - n} = \frac{ (n-1)N }{ N - n} = \frac{ n - 1}{ 1 - \frac{n}N }~,$$
the same as that for $E(\hat s^2)$ on the left hand side. This proves $E(\hat s^2) = V( \bar x^2 )$

Besides the algebraic manipulation similar to the above, a classic proof can be seen in chapter 2 of a well-known old text by Cochran. In particular, if you can get a copy, the author establishes a hierarchy of relations of great importance from Theorem 2.1 going to Theorem 2.4 while emphasizing the concepts instead of the algebra.
This might not be what you started out having in mind, but it provides a different flavor to say the least.
