Let $X_1,\ldots,X_n$ be independent $\mathcal{N}(\mu, \sigma^2).$
I got stumbled in one term of a task. I know
$$\mathbb E\big[\sum_{i=1}^{n} (X_i-\bar X)^2\big]= \sum_{i=1}^{n}\mathbb E\big[(X_i-\bar X)^2\big] =\sum_{i=1}^{n}\mathbb V(X_i)=n \sigma^2.$$
But how can I simplify the following term by an expression that only involved $n$ and $\sigma^2$?
$$\mathbb E\big[\big\{\sum_{i=1}^{n} (X_i-\bar X)^2\big\}^2\big]=?$$
I tried by the following: $$\mathbb E\big[\big\{\sum_{i=1}^{n} (X_i-\bar X)^2\big\}^2\big]$$ $$=\mathbb E\big[\big\{\sum_{i=1}^{n} (X_i^2-2X_i\bar X+\bar X^2)\big\}^2\big]$$ $$=\mathbb E\big[\big\{\sum_{i=1}^{n}X_i^2-n\bar X^2)\big\}^2\big]$$ $$=\mathbb E\big[\big(\sum_{i=1}^{n}X_i^2\big)^2-2n\bar X^2\sum_{i=1}^{n}X_i^2+n^2\bar X^4]$$
Then, I don't know how is to simplify it more.
Your first expression is not correct, the result is $(n-1)\sigma^2$. To see this, observe that the sample variance can be written as $(X-\mu)^\top M (X-\mu)$ where $M$ is an idempotent matrix with rank $n-1$. This also shows that the distribution of the sample variance is proportional to $\chi^2_{n-1}$. Now all you need is the mean and variance of $\chi^2$ distribution which one can find, for example, in Mood, Graybill, and Boes.