Prove that $\frac{\sum_{i=1}^n (X_i - \bar{X})^2}{\sigma^2}\,\sim \,\chi_{(n-1)}^2$ If $X_1, \ldots, X_n$ are iid $\sim N(\mu,\sigma^2)$,
$\frac 1 {\sigma^2} \Big((X_1-\bar X)^2 + \cdots + (X_n - \bar X)^2 \Big) \sim \chi^2_{n-1}$.
This has been shown with linear algebra projections, and is the starting point to derive the density of the $t$-Student distribution.
I would really like to get a proof without the advanced math in linear algebra projections (if possible), or with a detailed explanation of the steps followed in the derivation (if the use of projections is essential). Thanks!
 A: Observe the following:
Since $X_i \sim N(\mu, \sigma^2) \Rightarrow X_i - \mu \sim N(0, \sigma^2) $ and that $\frac{X_i - \mu}{\sigma} \sim N(0,1)$. 
Now, by definition, a $\chi^2_1$ random variable is defined as $Z^2$ where $Z \sim N(0,1)$. 
So that means $(\frac{X_i - \mu}{\sigma})^2 \sim N(0,1)^2 = \chi^2_1$.
Finally, observe that $\sum_{i=1}^n \chi^2_1 = \chi^2_n$. This can be seen from the proof here: 
https://onlinecourses.science.psu.edu/stat414/node/171
So now we have shown that $\sum_{i=1}^n (\frac{X_i - \mu}{\sigma})^2 \sim \chi^2_n$.
We're almost there. In your case you replace the $\mu$ with a $\bar{X}$. That's standard practice in statistics when you don't know the true value of the parameter and in that case, we reduce the degrees of freedom by one. For that same reason, we have that $\sum_{i=1}^n (\frac{X_i - \bar{X}}{\sigma})^2 \sim \chi^2_{n-1}$.
If you want a rigorous proof of why the minus one, then you will need to get into the linear algebra. A quick discussion can be found here:
https://en.wikipedia.org/wiki/Degrees_of_freedom_%28statistics%29
Hope that helps!
A: Here's a writeup of this exact problem I made a few months ago. This is only the special case for i.i.d $\mathsf{N}(0,1)$, but it should only be off by a few multiplicative constants from the general case.
Derivation of the Chi-Squared distribution
The seek to find the CDF of the Chi-Squared distribution, that is, a function such that for $n$ i.i.d normally distributed CRVs $Z_1,...,Z_n\sim\mathsf{N}(0,1)$
$$\Xi(\chi^2)=\Pr\left(\sum_{i=1}^n {Z_i}^2< \chi^2\right)$$To start we consider the random vector
$$\underline{Z}=(Z_1,...,Z_n)^{\mathrm{T}}$$Which is distributed according to the PDF
$$\varphi(\underline{z})= \frac{1}{\left(\sqrt{2\pi }\right)^n}\exp\left(\frac{-\Vert\underline{z}\Vert^2}{2}\right)$$So then
$$\Pr\left(\Vert\underline{Z}\Vert^2<\chi^2\right)=\int\limits_{\mathcal{B}(0,\chi)}\varphi(\underline{z})\mathrm{d}V$$To do this we assume $n>3$ and convert to hyperspherical coordinates. The $n=1,2$ cases are easy and can be worked through individually.$$z_1=r\cos(\theta_1)~;~z_n=r\prod_{i=1}^{n-1}\sin(\theta_i)$$And for $k\notin \{1,n\}$
$$z_k=r\cos(\theta_k)\prod_{i=1}^{k-1}\sin(\theta_i)$$In this coordinate system the volume element is
$$\mathrm{d}V=r^{n-1}\mathrm{d}r\mathrm{d}\theta_{n-1}\prod_{k=1}^{n-2}\sin^{n-k-1}(\theta_k)\mathrm{d}\theta_k$$Then our integral is
$$\frac{1}{\left(\sqrt{2\pi}\right)^n}\underbrace{\int\limits_0^\pi \cdots\int\limits_0^\pi}_{n-2\text{ of these}}\int\limits_0^{2\pi} \int\limits_0^\chi \exp\left(\frac{-r^2}{2}\right)r^{n-1}\mathrm{d}r\mathrm{d}\theta_{n-1}\prod_{k=1}^{n-2}\sin^{n-k-1}(\theta_k)\mathrm{d}\theta_k$$We can express the $r$ integral in terms of the lower incomplete Gamma function
$$\int_0^\chi \exp\left(\frac{-r^2}{2}\right)r^{n-1}\mathrm{d}r=\left(\sqrt{2}\right)^{n-2}\gamma\left(\frac{n}{2},\frac{\chi^2}{2}\right)$$Hence after integrating with respect to $\theta_{n-1}$
$$\Pr(\Vert\underline{Z}\Vert^2<\chi^2)=\frac{\gamma\left(\frac{n}{2},\frac{\chi^2}{2}\right)}{\sqrt{\pi}^{n-2}} \int\limits_0^\pi \cdots\int\limits_0^\pi\prod_{k=1}^{n-2}\sin^{n-k-1}(\theta_k)\mathrm{d}\theta_k$$From here we make use of the integral identity$$\int_0^\pi \sin^m(x)\mathrm{d}x=\sqrt{\pi}\frac{\Gamma\left(\frac{m}{2}+\frac{1}{2}\right)}{\Gamma\left(\frac{m }{2}+1\right)}\tag1{}$$So
$$\Pr(\Vert\underline{Z}\Vert^2<\chi^2)=\gamma\left(\frac{n}{2},\frac{\chi^2}{2}\right)\prod_{k=1}^{n-2}\frac{\Gamma\left(\frac{n-k}{2}\right)}{\Gamma\left(\frac{n-k+1}{2}\right)}\tag{2}$$The above product (2) simplifies quite nicely, and we get the chi-squared CDF
$$\Xi(x;n):=\Pr(\Vert\underline{Z}\Vert^2<x)=\frac{\gamma\left(\frac{n}{2},\frac{x}{2}\right)}{\Gamma(n/2)}$$Which we can differentiate to get the chi squared PDF:
$$\xi(x;n)=\frac{x^{n/2-1}e^{-x/2}}{2^{n/2}\Gamma(n/2)}$$

PROOFS OF IDENTITIES (1),(2)
We'll start with the integral -
$$I_m=\int_0^\pi \sin^m(x)\mathrm{d}x$$Integrate by parts. Let $u=\sin^{m-1}(x)$, $\mathrm{d}v=\sin(x)\mathrm{d}x$ $$I_m=\int_0^\pi u~\mathrm{d}v=\underbrace{(uv)\big|^\pi_0}_{\sin0=\sin\pi=0}-\int_0^\pi v~\mathrm{d}u$$$$I_m=(m-1)\int_0^\pi\cos^2(x)\sin^{m-2}(x)\mathrm{d}x$$$$\frac{I_m}{m-1}=\int_0^\pi \sin^{m-2}(x)\mathrm{d}x-\int_0^\pi \sin^m(x)\mathrm{d}x$$So we have a recurrence relation$$\frac{I_m}{m-1}+I_m=I_{m-2}\text{ or, equivalently }I_{m+2}=\frac{m+1}{m+2}I_m$$With initial data $I_0=\pi=\Gamma(1/2)^2$, $I_1=2$. One can check that$$I_m=\sqrt{\pi}\frac{\Gamma\left(\frac{m+1}{2}\right)}{\Gamma\left(\frac{m}{2}+1\right)}$$Satisfies the recurrence. You also could get the solution constructively by using the recursive property of Gamma, namely, $\Gamma(z+1)=z\Gamma(z)$ on the denominator and the Legendre duplication formula on the numerator:$$\Gamma(2z)=\frac{2^{2z-1}\Gamma(z)\Gamma(z+1/2)}{\sqrt{\pi}}$$This gives that for integer $m$,
$$\Gamma\left(\frac{m}{2}+\frac{1}{2}\right)=\frac{\sqrt{\pi}(m-1)!}{2^{m-1}\Gamma(m/2)}$$So one could write the solution as $$I_m=\frac{\pi}{2^{m-2}m}\frac{(m-1)!}{\Gamma(m/2)^2}~~|~~m\neq0$$Perhaps this is more readily obtainable given our initial conditions.
Now the proof of the product identity,
$$P_n=\prod_{k=1}^{n-2}\frac{\Gamma\left(\frac{n-k}{2}\right)}{\Gamma\left(\frac{n-k+1}{2}\right)}=\frac{1}{\Gamma(n/2)}$$ We can do some index shifting:$$\prod_{k=1}^{n-2}\frac{\Gamma\left(\frac{n-k}{2}\right)}{\Gamma\left(\frac{n-k+1}{2}\right)}=\prod_{k=2}^{n-1}\frac{\Gamma(k/2)}{\Gamma\left(\frac{k+1}{2}\right)}$$We can index shift again to
$$P_n=\prod_{k=1}^{n-2}\frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k+2}{2}\right)}=\frac{\prod_{i=2}^{n-1}\Gamma(i/2)}{\prod_{j=3}^{n}\Gamma(j/2)}$$Expanding,
$$P_n=\frac{\Gamma(2/2)\Gamma(3/2)...\Gamma((n-1)/2)}{\Gamma(3/2)...\Gamma((n-1)/2)\Gamma(n/2)}$$Hence
$$P_n=\frac{1}{\Gamma(n/2)}.$$
