Show that $\sqrt n(G_n − \frac1e) →^d N(0,\sigma^2)$

Question

Let $X_1,X_2\ldots $ be a sequence of independent, identically distributed with $X_i\sim U(0,1)$

For the sequence of geometric means $G_n = \big(\prod_{i=1}^n X_i \big)^{1/n}$ show that $\sqrt n(G_n − \frac1e) →^d N(0,\sigma^2)$, for some $\sigma. $ Find $\sigma$

My attempt:

$\ln(G_n)= \frac1n \ln(\prod_{i=1}^n X_i) = \frac1n \sum_1^n \ln(X_i)$

Let $Y_i = \ln(X_i) \Rightarrow \ln(G_n) = \overline{Y} = \frac1n \sum_1^n Y_i$

I found out that, $ \mu = E(Y_1) = \int_0^1 \ln x \;dx = -1$ and $\sigma^2 = Var(Y_1) = \int_0^1 (\ln(x) +1)^2 \; dx = 1$

I know by Central Limit Theorem, that

$$\frac{\sqrt n (\overline{Y} - (-1))}{1} \sim N(0,1) $$

$$ \Rightarrow \sqrt n (\overline{Y} - (-1)) \sim N(0,1) $$

Using delta method with $f(x) = e^x$, we get

$$ \Rightarrow \sqrt n (e^{\overline{Y}} - \frac1e) \sim \frac1e N(0,1) $$

$$ \Rightarrow \sqrt n (G_n - \frac1e) \sim \frac1e N(0,1) \sim N(0,\frac1{e^2})$$

So, $ \sigma = \frac1e$

Is this correct?