= Solution
The <likelihood> is $L(\theta;x)=\prod_{i=1}^nf(x_i;\theta)$, and a <maximum> <likelihood> estimator is any maximizer $\hat\theta\in\arg\max_\theta L(\theta;X)$. Write $\ell=\log L$ and score $U=\nabla_\theta\ell$. <Fisher information> for the sample is
$$
I_n(\theta)=\mathbb E_\theta[UU^T]=-\mathbb E_\theta[\nabla_\theta^2\ell],
$$
where the equality uses the usual <differentiability> and interchange assumptions. For <independent> identically distributed observations $I_n=nI_1$. Under standard interior, identifiability and regularity assumptions,
$$
\boxed{\sqrt n(\hat\theta-\theta)\Rightarrow N_d(0,I_1(\theta)^{-1}),\quad\hat\theta\approx N_d(\theta,I_n(\theta)^{-1}).}
$$
Without those assumptions the large-sample normal assertion need not hold, for example at a boundary parameter.
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