Problem understanding the theorem

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:

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

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

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

since $E[(X_i - \mu)^2] = \sigma^2$ and $E[(\bar X - \mu)^2] = \frac{\sigma^2}{n}$ it follows that
$<br /> <br /> E(S^2) = \frac{1}{n-1} \left[\sum_{i=1}^{n} \sigma^2 - n \frac{\sigma^2}{n}\right]$ ...............Understood!!!!

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

How did he get $\sigma^2$