Proof that the sampling distribution of the sample variance from $N(0,1)\sim \chi_{n-1}^2$ 
Is this true? How to verify it? 
From the definition of chi square I can not judge whether it is chi square.
 A: Let $X_{i}$ for $i=1,2,\cdots n$ are independent $N(0,1)$ random variables.
Then,  the distribution of $\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}\sim \chi_{n-1}^{2}$.\
Let $\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}=(n-1)S^{2}$. 
\begin{eqnarray}
\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}&=&\underbrace{\sum_{i=1}^{n} X_{i}^{2}}_{\chi_{n}^{2}}-n\bar{X}^{2}\nonumber\\
(n-1)S^{2}&=&\sum_{i=1}^{n} X_{i}^{2}-n\bar{X}^{2}\nonumber\\
\sum_{i=1}^{n} X_{i}^{2}&=& (n-1)S^{2} + n\bar{X}^{2}
\end{eqnarray}
Now consider,
\begin{eqnarray*}
X_{i}\sim N(0,1)\\
\bar{X}\sim N(0,\frac{1}{n})\\
\frac{\bar{X}}{1/\sqrt{n}}\sim N(0,1)\\
n\bar{X}^{2}\sim \chi_{1}^{2}\\
\end{eqnarray*}
Identifying the distribution of the terms in the earlier equation, it can be expressed as
\begin{equation}
\chi_{n}^{2} = (n-1)S^{2}+\chi_{1}^{2}
\end{equation}
By a property of chi-squared distribution, $S^{2}$ and $\bar{X}$ are independent. That is, right hand side of the above equation is sum of two independent random variables. Now consider the moment generating function on both sides. The left hand side random variable being a $\chi_{n}^{2}$ random variable, its MGF is $(1-2t)^{-n/2}$. 
Substituting the MGF values and simplifying, 
\begin{eqnarray*}
M_{\chi_{n}^{2}}(t) = M_{(n-1)S^{2}}(t)\cdot M_{\chi_{1}^{2}}(t)\\
(1-2t)^{-n/2} = M_{(n-1)S^{2}}(t) \cdot (1-2t)^{-1/2}\\
M_{(n-1)S^{2}}(t) = (1-2t)^{-n/2} \cdot  (1-2t)^{-1/2}\\
M_{(n-1)S^{2}}(t) = (1-2t)^{-(n-1)/2}
\end{eqnarray*}
which is the MGF of a $\chi_{n-1}^{2}$ random variable. Hence, 
 $(n-1)S^{2}\sim\chi_{n-1}^{2}$.
