How come this is Chi-Squared distributed? Let  $X_1, X_2, ..., X_n$ be normally distributed with mean $\mu$ and variance $\sigma ^2$, then $\frac{(n-1) S^2}{ \sigma^2}$ has a Chi-Square distribution with $n-1$ degrees of freedom. 
How come it's chi-square distributed?
Attempt:
$S^2 = \frac{1}{n-1} \sum(X_i-\overline X)^2 = \frac{1}{n-1}(\sum X_i ^2 - (\sum X_i)^2)$.  
Here, the first term $\sum X_i^2$ is chi-squared. But you also need to subtract the second term (which is normal distributed with mean $\mu$, and variance $\frac{\sigma^2}{n}$. How does it make it chi-square distributed?
 A: The first proof people generally encounter involves MGFs. Note, 
$$ \frac{(n-1)S^2}{\sigma^2} = \sum\limits_i^n\left(\frac{X_i-\bar{X}}{\sigma}\right)^2 = \sum\limits_i^n\left(\frac{X_i-\mu}{\sigma}\right)^2 - \left(\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\right)^2$$
Observe that the first term on the RHS is a chi-square RV with $n$ degrees of freedom. The second term is a chi-square with $1$ degree of freedom. Rearranging,
$$ \frac{(n-1)S^2}{\sigma^2} + \left(\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\right)^2 = \sum\limits_i^n\left(\frac{X_i-\mu}{\sigma}\right)^2 $$
Now we can find the MGFs both sides, 
$$ (1-2t)^{-1/2}M(t) = (1-2t)^{-n/2} \Longrightarrow M(t) = (1-2t)^{-n/2+1/2} = (1-2t)^{-1/2(n-1)}$$
where $M(t)$ is the MGF of $\frac{(n-1)S^2}{\sigma^2}$. Thus, the MGF of $\frac{(n-1)S^2}{\sigma^2}$ is that of a chi-square RV with $n-1$ degrees of freedom.
Note: The above used the identity that $M_{X+Y}(t) = M_X(t)M_Y(t)$ where $X$ and $Y$ are independent. This detail was skipped, but one also needs to show that $S^2$ and $\bar{X}$ are independent which is true in the case of a Normally distributed sample. 
