Given a set of i.i.d sampled points $(A_1, B_1), \ldots, (A_n, B_n)$ and the correlation coefficient $\rho = Corr(A_1, B_1)$. Define $\hat{\rho}$ as the method-of -moments estimator of $\rho$. Show that $\sqrt{n}(\hat{\rho} - \rho)\rightarrow N(0, (1-\rho^2)^2)$ in distribution asymptotically.
My attempt: First, it's easy to see that $\hat{\rho} = \frac{\frac{1}{n}\sum_{i=1}^{n} A_iB_i - (\frac{1}{n}\sum_{i=1}^{n} A_i)(\frac{1}{n}\sum_{i=1}^{n} B_i)}{\sqrt{\frac{1}{n-1}\sum_{i=1}^{n} (A_i - \overline{A})^2}\sqrt{\frac{1}{n-1}\sum_{i=1}^{n} (B_i - \overline{B})^2}}$. Now, by using Lindeberg-Levy's Central Limit Theorem, we have:
$\sqrt{n}(\frac{1}{n}\sum_{i=1}^{n} A_iB_i - E(A_1B_1))\rightarrow N(0,\sigma_{A_1B_1}^{2})$ in distribution $\ \ \ (1)$
Also, by Strong Law of Large Number and $A_1,\ldots, A_n$, $B_1,\ldots, B_n$ are i.i.d, we have:
$\sqrt{\frac{1}{n-1}\sum_{i=1}^{n} (A_i - \overline{A})^2}\rightarrow \sigma_{A_1}$ almost surely.
$\sqrt{\frac{1}{n-1}\sum_{i=1}^{n} (B_i - \overline{B})^2}\rightarrow \sigma_{B_1}$ almost surely.
Now, if we rewrite $\sqrt{n}[-(\frac{1}{n}\sum_{i=1}^{n} A_i)(\frac{1}{n}\sum_{i=1}^{n} B_i) + E(A_1)E(B_1)] = \frac{1}{\sqrt{n}}[\sqrt{n}(-(\frac{1}{n}\sum_{i=1}^{n} A_i-E(A_1))\sqrt{n}((\frac{1}{n}\sum_{i=1}^{n} B_i-E(B_1)))] + \sqrt{n}[-E(B_1)(\frac{1}{n}\sum_{i=1}^{n} A_i)-E(A_1)(\frac{1}{n}\sum_{i=1}^{n} B_i)+2E(A_1)E(B_1)]$,
then combining Lindeberg-Levy's Central Limit Theorem with Strong Law of Large Number gives us:
$$\sqrt{n}[-E(B_1)(\frac{1}{n}\sum_{i=1}^{n} A_i)-E(A_1)(\frac{1}{n}\sum_{i=1}^{n} B_i) + 2E(A_1)E(B_1)]\rightarrow -E(B_1)E(A_1) - E(A_1)E(B_1) + 2E(A_1)E(B_1) =0$$
$$\frac{1}{\sqrt{n}}[\sqrt{n}(-(\frac{1}{n}\sum_{i=1}^{n} A_i-E(A_1))\sqrt{n}(-(\frac{1}{n}\sum_{i=1}^{n} B_i-E(B_1)))]\rightarrow \frac{1}{\sqrt{n}} N(0, \sigma_{A_1}^{2})N(0,\sigma_{B_1}^2)\ \ \ (2)$$
From $(1)$ and $(2)$, we obtain: $\sqrt{n}(\hat{\rho} - \rho)\rightarrow N(0, \sigma_{A_1B_1}^{2}) - \frac{1}{\sqrt{n}}N(0, \sigma_{A_1}^{2})N(0,\sigma_{B_1}^2)$.
My question: Can anyone please give some thoughts on how to show that the variance of
$$N(0, \sigma_{A_1B_1}^{2}) - \frac{1}{\sqrt{n}}N(0, \sigma_{A_1}^{2})N(0,\sigma_{B_1}^2)$$
equals to $$(1-\rho^2)^2$$