Theorem : Is S^2 is the variance of a random sample from a infinite population with the finite variance \sigma^2 , then E(S^2) = \sigma^2

Proof:

$\displaystyle E(S^2) = E \left[ \frac{1}{n-1} . \sum_{i=1}^{n} (X_i - \bar X)^2 \right]$..............Understood!!!!

$\displaystyle E(S^2) = \frac{1}{n-1} . E \left[\sum_{i=1}^{n} [(X_i - \mu) - (\bar X - \mu)]^2 \right]$.............Understood!!!!

$\displaystyle E(S^2) = \frac{1}{n-1} . \left[\sum_{i=1}^{n} E[(X_i - \mu)^2] - n E[(\bar X - \mu)^2] \right]$...................Understood!!!!

since $\displaystyle E[(X_i - \mu)^2] = \sigma^2$ and $\displaystyle E[(\bar X - \mu)^2] = \frac{\sigma^2}{n}$ it follows that

$\displaystyle

E(S^2) = \frac{1}{n-1} \left[\sum_{i=1}^{n} \sigma^2 - n \frac{\sigma^2}{n}\right]$...............Understood!!!!

$\displaystyle = \sigma^2$....................Couldnt$\displaystyle \sigma^2$Understood!!!!

How did he get