Prove $\operatorname{Cov}(\overline{X_n}, X_j - \overline{X_n}) = 0$ for independent normally distributed random variables My homework states the following problem:

Let $X_1, \dots, X_n$ be independent $N(\mu, \sigma^2)$ distributed random variables, $\overline{X_n}$ be the sample mean and $S_n^2$ the empirical variance. Show that $\operatorname{Cov}(\overline{X_n}, X_j - \overline{X_n}) = 0$ and conclude that $\overline{X_n}$ and $S_n^2$ are independent.

My first approach:
$$ \begin{align*}
  \operatorname{Cov}(\overline{X_n}, X_j - \overline{X_n})
    &= \mathbb E(\overline{X_n} X_j - \overline{X_n}^2) - \mathbb E(\overline{X_n}) \mathbb E(X_j - \overline{X_n}) \\
    &= \mathbb E(\overline{X_n} X_j) - \mathbb E(\overline{X_n}^2) - \mathbb E(\overline{X_n}) \mathbb E(X_j - \overline{X_n}) \\
\end{align*} $$
My second approach is
$$ \begin{align*}
  \operatorname{Cov}(\overline{X_n}, X_j - \overline{X_n})
    &= \mathbb E\left[(\overline{X_n} - \mathbb E(\overline{X_n})) (X_j - \overline{X_n} - \mathbb E(X_j - \overline{X_n}))\right]
\end{align*} $$
I always end up with expressions involving the sample mean and the population mean. My problem is: I know about the Law of Large Numbers bridging the gap between samples and populations, but here no series is given, but finite $n$. Which theorem can help me to solve this problem?
Update: A solution to the second question is given at jekyll.math.byuh.edu
 A: You have \begin{align}\text{Cov}(\bar X_n,X_j-\bar X_n)&=\text{Cov}(X_j,\bar X_n)-\text{Var}(\bar X_n)\\&=\text{Cov}\left(X_j,\frac{1}{n}\sum_{i=1}^nX_i\right)-\text{Var}(\bar X_n)\\&=\frac{1}{n}\sum_{i=1}^n\text{Cov}(X_i,X_j)-\text{Var}(\bar X_n)\\&=\frac{1}{n}\left(\text{Var}(X_i)+\sum_{i\ne j}\text{Cov}(X_i,X_j)\right)-\text{Var}(\bar X_n)\\&=\frac{1}{n}\text{Var}(X_i)-\text{Var}(\bar X_n)\\&=\frac{\sigma^2}{n}-\frac{\sigma^2}{n}=0\end{align}
Now you have to prove that $(\bar X_n,X_j-\bar X_n)$ is jointly normal for all $j=1,2,\cdots,n$ using MGF or otherwise. Once you have proved the joint normality, then the fact that $\bar X_n$ and $X_j-\bar X_n$ are uncorrelated would imply their independence. That is, $\bar X_n$ is independent of $X_1-\bar X_n,X_2-\bar X_n,\cdots,X_n-\bar X_n$, and hence also independent of $S_n^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X_n)^2$.
To show the joint normality of $\bar X_n$ and $X_j-\bar X_n$, note that both $\bar X_n$ and $X_j-\bar X_n$ are linear combinations of independent normal variables $X_1,X_2,\cdots,X_n$ for all $j=1,2,\cdots,n$. As such, their joint distribution $(\bar X_n,X_j-\bar X_n)$ has to be bivariate normal. 
For a formal proof, you may find the joint moment generating function (MGF) of, say, $(\bar X_n,X_1-\bar X_n)$ and show that the MGF is the MGF of a bivariate normal distribution.The details using MGF might get complicated, but to use the zero covariance proved in the first part to finally prove the independence of $(\bar X_n,S_n^2)$ you would have to show the joint normality somehow.
