given a random sample with the following density $$e^{\theta-xe^{\theta}}$$ Find maximum likelihood estimator for $\theta$ e specify the asymptotic distribution.
I useded $\lambda = e^{\theta}$ $\theta = log(\lambda)$ in order to recall the exponential distribution I showed that $\hat {\theta}_{MLE}= -log(\bar X)$ because $\hat {\lambda}_{MLE}= \frac{1}{\bar X}$
how can I specify the asymptotic variance of the $\hat {\theta}_{MLE}$? I know the asymptotic variance it's the inverse of the fisher information for the exponential distribution the asymptotic variance of the MLE is given by this expression $\frac{\lambda_0^2}{n}$.by substituting $\lambda = e^{\theta}$ I get that the asymptotic variance is equal to $\frac{e^{2\theta}}{n}$ . the solution should be $\frac{1}{n}$ for the asymptotic variance of $\hat {\theta}_{mle}$ , I do not understand why
can someone give an hint?