How to prove $s^2$ is a consistent estimator of $\sigma^2$? I am trying to prove that $s^2=\frac{1}{n-1}\sum^{n}_{i=1}(X_i-\bar{X})^2$ is a consistent estimator of $\sigma^2$ (variance), meaning that as the sample size $n$ approaches $\infty$ , $\text{var}(s^2)$ approaches 0 and it is unbiased.
I understand how to prove that it is unbiased, but I cannot think of a way to prove that $\text{var}(s^2)$ has a denominator of n. Does anyone have any ways to prove this?

The question asks:
A random sample of size n is taken from a normal population with variance $\sigma^2$. Show that the statistic $s^2$ is a consistent estimator of $\sigma^2$
So far I have gotten: 
$\text{var}(s^2) = \text{var}(\frac{1}{n-1}\Sigma X^2-n\bar X^2)$ 
$= \frac{1}{(n-1)^2}(\text{var}(\Sigma X^2) + \text{var}(n\bar X^2))$ 
$= \frac{n^2}{(n-1)^2}(\text{var}(X^2) + \text{var}(\bar X^2))$
But as I do not know how to find $Var(X^2) $and$ Var(\bar X^2)$, I am stuck here (I have already proved that $S^2$ is an unbiased estimator of $Var(\sigma^2)$)
Source : Edexcel AS and A Level Modular Mathematics S4 (from 2008 syllabus) Examination Style Paper Question 1. This is for my own studies and not school work.
 A: It's a very well known result that :

If $X_1, X_2, \cdots, X_n \stackrel{\text{iid}}{\sim} N(\mu,\sigma^2)$ , then $$Z_n = \dfrac{\displaystyle\sum(X_i - \bar{X})^2}{\sigma^2} \sim \chi^2_{n-1}$$
Thus, $ \mathbb{E}(Z_n) = n-1 $  and  $ \text{var}(Z_n) = 2(n-1)$ .

If you wish to see a proof of the above result, please refer to this link.
Now, since you already know that $s^2$ is an unbiased estimator of $\sigma^2$ , so for any $\varepsilon>0$ , we have :
\begin{align*}
&\mathbb{P}(\mid s^2 - \sigma^2 \mid > \varepsilon )\\
&= \mathbb{P}(\mid s^2 - \mathbb{E}(s^2) \mid > \varepsilon )\\
&\leqslant \dfrac{\text{var}(s^2)}{\varepsilon^2}\\
&=\dfrac{1}{(n-1)^2}\cdot \text{var}\left[\sum (X_i - \overline{X})^2)\right]\\
&=\dfrac{\sigma^4}{(n-1)^2}\cdot \text{var}\left[\frac{\sum (X_i - \overline{X})^2}{\sigma^2}\right]\\
&=\dfrac{\sigma^4}{(n-1)^2}\cdot\text{var}(Z_n)\\
&=\dfrac{\sigma^4}{(n-1)^2}\cdot 2(n-1) = \dfrac{2\sigma^4}{n-1} \stackrel{n\to\infty}{\longrightarrow} 0
\end{align*}
Thus, $ \displaystyle\lim_{n\to\infty} \mathbb{P}(\mid s^2 - \sigma^2 \mid > \varepsilon ) = 0$ , i.e. $ s^2 \stackrel{\mathbb{P}}{\longrightarrow} \sigma^2 $ as $n\to\infty$ , which tells us that $s^2$ is a consistent estimator of $\sigma^2$ .

Note : I have used Chebyshev's inequality in the first inequality step used above. Hope my answer serves your purpose. Thank you.
