The MLE for a collection of $n$ Poisson random variables can be determined with the standard method; set the derivative of the log-likelihood to $0$, and what pops out is that $$\hat{\lambda} = \frac{1}{n}\sum_{i=1}^nX_i.$$ My question is about the asymptotic normality of this MLE, with asymptotic normality defined by: $$\sqrt{n} (\hat{\theta}_n - \theta_0) \to^d N\left(0, \frac{1}{I(\theta_0)}\right),$$ where $\theta_0$ is the true parameter and $I$ is the Fisher information.
Now, to determine the asymptotic normality, we first evaluate the Fisher information. In this case, $$I(\lambda) = \frac{n}{\lambda},$$ so $$\sqrt{n} (\hat{\lambda} - \lambda) \to^d N\left(0, \frac{\lambda}{n}\right).$$
I would assume that this means that $$(\hat{\lambda} - \lambda) \to^d N\left(0, \frac{\lambda}{n^2}\right),$$ and thus $$\hat{\lambda} \to^d N\left(\lambda, \frac{\lambda}{n^2}\right).$$
However, it turns out that this is wrong: it should be $$\hat{\lambda} \to^d N\left(\lambda, \frac{\lambda}{n}\right),$$ but I can't figure out why. Where did I go wrong?