Prove the sample variance is an unbiased estimator I have to prove that the sample variance is an unbiased estimator. What is is asked exactly is to show that following estimator of the sample variance is unbiased:

$s^2=\frac{1}{n-1}\sum\limits_{i=1}^n(x_i-\bar x)^2$

I already tried to find the answer myself, however I did not manage to find a complete proof.
For a shorter proof, here are a few things we need to know before we start:

$X_1, X_2 , ..., X_n$ are independent observations from a population with mean $\mu$ and variance $\sigma^{2}$

$\mathbb E(X_i) = \mu$ , $\mathbb{Var}(X_i)= \sigma^{2}$

$\mathbb E(X^2) = \sigma^{2} + \mu^{2}$

$\mathbb{Var}(X)=\mathbb E(X^2)-\mathbb [E(X)]^2$

$\mathbb E(\bar{X}^2) = \frac{\sigma^2}{n} + \mu^2$



Let's try to show that $\mathbb E(s^2)= \mathbb E\left(\frac{\sum_{i=1}^n (X_i - \bar X)^2}{n-1}\right) = \sigma^{2}$

To make my life easier, I will omit the limits of summation from now onwards, but let it be known that we are always summing from $1$ to $n$.

$\mathbb E\left(\sum (X_i - \bar X)^2 \right) = \mathbb E\left(\sum X_{i}^2 - 2 \bar X \sum X_i + n \bar X^2 \right) = \sum \mathbb E(X_{i}^2) - \mathbb E\left(n \bar X^2 \right)$

$\sum \mathbb E(X_{i}^2) - \mathbb E\left(n \bar X^2 \right) = \sum \mathbb E(X_{i}^2) - n \mathbb E\left(\bar X^2\right) = n \sigma^2 + n \mu^2 - \sigma^2 -n \mu^2$

This simplifies to $(n-1) \sigma^2$

So far, we have shown that $\mathbb E\left(\sum (X_i - \bar X)^2 \right) = (n-1)\sigma^2$

$\mathbb E(s^2)= \mathbb E\left(\frac{\sum (X_i - \bar X)^2}{n-1}\right) = \frac{1}{n-1} \mathbb E\left(\sum (X_i - \bar X)^2 \right)$

$\mathbb E(s^2) = \frac {(n-1)\sigma^2}{n-1} = \sigma^2$

We have now shown that the sample variance is an unbiased estimator of the population variance.