Let $X$ have an exponential distribution with parameter $\theta$ (pdf is $f(x, \theta) = \theta e^{-\theta x}$). I already found that the MLE for $\theta$ after $n$ observations is $$\hat{\theta}_{MLE} = \bar{X}^{-1} = \frac{n}{\sum_{i=1}^n{X_i}}$$ and that $\bar{X} \tilde{} \Gamma(n, n\theta)$.

The question is to derive **directly** (i.e. without using the general theory for asymptotic behaviour of MLEs) the asymptotic distribution of $$\sqrt n (\hat{\theta}_{MLE} - \theta)$$
as $n \to \infty$.

According to the general theory (which I should not be using), I am supposed to find that it is asymptotically $N(0, I(\theta)^{-1}) = N(0, \theta^2)$. However, I don't know where to start - for other distributions I was able to use CLT (if their MLE was the sample mean), but I can't think of a way to do it here.