Observed Fisher information for the binomial: How is $I(\hat{\theta}) = \frac{n}{\hat{\theta}(1 - \hat{\theta})}$ calculated? I am currently studying the textbook In All Likelihood by Yudi Pawitan. Example 2.10 of chapter 2.5 Maximum and curvature of likelihood says the following:

Example 2.10: Based on $x$ from the binomial$(n, \theta)$ the log-likelihood function is
$$\log L(\theta) = x \log \theta + (n - x) \log(1 - \theta).$$
We can first find the score function
$$\begin{align} S(\theta) &\equiv \frac{\partial}{\partial{\theta}} \log L(\theta) \\ &= \frac{x}{\theta} - \frac{n - x}{1 - \theta}, \end{align}$$
giving the MLE $\hat{\theta} = x/n$ and
$$\begin{align} I(\theta) &\equiv -\frac{\partial^2}{\partial{\theta}^2} \log L(\theta) \\ &= \frac{x}{\theta^2} + \frac{n - x}{(1 - \theta)^2}, \end{align}$$
so at the MLE we have the Fisher information
$$I(\hat{\theta}) = \frac{n}{\hat{\theta}(1 - \hat{\theta})}. \square$$

How is $I(\hat{\theta}) = \frac{n}{\hat{\theta}(1 - \hat{\theta})}$ calculated? I know that the MLE $\hat{\theta}$ is the solution of the score equation $S(\theta) \equiv \frac{\partial}{\partial{\theta}} \log L(\theta) = 0$, and I know that $I(\theta) \equiv - \frac{\partial^2}{\partial{\theta}^2} \log L(\theta)$ is the Fisher information, but I can't figure out precisely what $I(\hat{\theta})$ is, or exactly how $I(\hat{\theta}) = \frac{n}{\hat{\theta}(1 - \hat{\theta})}$ was calculated.

Hint: Substitute the value $\hat\theta=x/n$ in $I(\theta) .$ See what happens.
Substitute in place of $\theta$ to get $$\frac{n^2x}{x^2}+ \frac{n^2(n-x) }{(n-x) ^2}.$$ Simplify it.
