Since the $X_i$ are i.i.d., the joint density factorises: $f(x_1, \ldots, x_n, \theta) = \prod_{i=1}^n f(x_i, \theta)$. Taking logarithms gives \begin{align*} \log f(X_1, \ldots, X_n, \theta) = \sum_{i=1}^n \log f(X_i, \theta), \end{align*} so the score of the full sample is the sum of the individual scores: \begin{align*} \nabla_\theta \log f(X_1, \ldots, X_n, \theta) = \sum_{i=1}^n \nabla_\theta \log f(X_i, \theta). \end{align*} Expanding the outer product in the definition of $I_n(\theta)$: \begin{align*} I_n(\theta) = \mathbb{E}_\theta\!\left[\sum_{i=1}^n \sum_{j=1}^n \nabla_\theta \log f(X_i, \theta)\, \nabla_\theta \log f(X_j, \theta)^\top\right]. \end{align*} For $i \neq j$, independence and the fact that $\mathbb{E}_\theta[\nabla_\theta \log f(X_i, \theta)] = 0$ give \begin{align*} \mathbb{E}_\theta\!\left[\nabla_\theta \log f(X_i,\theta)\, \nabla_\theta \log f(X_j, \theta)^\top\right] = \mathbb{E}_\theta[\nabla_\theta \log f(X_i,\theta)]\, \mathbb{E}_\theta[\nabla_\theta \log f(X_j,\theta)]^\top = 0. \end{align*} Only the $n$ diagonal terms $i = j$ survive, each contributing $I(\theta)$ by the identical distribution of the $X_i$. Hence $I_n(\theta) = n\, I(\theta)$.