Maximum likelihood estimator and asymptotic distribution 
Let $X_1,\dots,X_n$ be a random sample from X whose density is given
by
$$f(x,\theta) = c(\theta)(1-\exp(-|x|))I\{|x|\leq\theta\}$$
Find the maximun likelihood estimator of $\theta$ and show that
$n(\hat{\theta}-\theta)$ converges in distribution to a gamma
distribution.

First we need to find $c(\theta)$, it's clear that it's a normalization constant, so integrating the density we get that $c(\theta)= 1/2(\theta+e^{-\theta}-1)$ and we may show that
\begin{align}
\mathcal{L}(\theta;x) &= \prod_{i=1}^n c(\theta)(1-\exp(-|x_i|))I\{|x_i|\leq\theta\} \\
 &=c^n(\theta)I_{(0,\theta)}\left(\frac{x_{(n)}-x_{(1)}}{2}\right) \prod_{i=1}^n (1-\exp(-|x_i|)) \\
 &=c^n(\theta)I\left(\frac{x_{(n)}-x_{(1)}}{2},+\infty\right)(\theta) \prod_{i=1}^n (1-\exp(-|x_i|))
\end{align}
and as $c(\theta)$ is a decreasing function of $\theta$, we get that
$$
\hat{\theta} =\frac{x_{(n)}-x_{(1)}}{2}
$$
however I'm stuck at the convergence part, I tried manipulating the expression and applying the Jacobian method. Jacobian yields a quite difficult convolution.
Other post shows something similar however the $X$s are exponentially distributed and the results follow from the memoryless property.
 A: \begin{align}
\mathcal{L}(\theta) &= \prod_{i=1}^n c(\theta)(1-\exp(-|X_i|))1\{|X_i|\leq\theta\} \\
 &=c^n(\theta)1\!\left\{\max_{1\le i\le n}|X_i|\le \theta\right\} \prod_{i=1}^n (1-\exp(-|X_i|)).
\end{align}
Since $c(\theta)$ is decreasing in $\theta$ but the indicator is $0$ when $\max_{1\le i\le n}|X_i|>\theta$, the MLE of $\theta$ is
$$
\hat{\theta}_n=\max_{1\le i\le n}|X_i|.
$$
Now, for $0<t<n\theta$ (note that $0\le \hat{\theta}_n\le \theta$),
\begin{align}
\mathsf{P}\!\left(n(\hat{\theta}_n-\theta)\le -t\right)&=\mathsf{P}\!\left(\max_{1\le i\le n}|X_i|\le \theta-\frac{t}{n}\right)=\left[\mathsf{P}\!\left(|X_1|\le \theta-\frac{t}{n}\right)\right]^n \\
&=\left[\left(e^{-\theta+\frac{t}{n}}+\theta-\frac{t}{n}-1\right)\left(\theta+e^{-\theta}-1\right)^{-1}\right]^n \\
&\to \exp\left(-\frac{t\left(e^{\theta}-1\right)}{e^{\theta}(\theta -1)+1}\right)
\end{align}
as $n\to\infty$. Thus, the distribution of $n(\theta-\hat{\theta}_n)$ converges to the gamma distribution with shape parameter $1$ and scale parameter:
$$
\frac{e^{\theta}(\theta-1)+1}{e^{\theta}-1}.
$$
