Additional comments: Your answer seems OK. It may be of interest to know that
$\hat \theta$ is not unbiased. One can get a rough idea of the distribution
of $\hat \theta$ for a particular $\theta$ by simulating many samples of
size $n.$ I don't know of a convenient 'unbiasing' constant multiple.
The Wikipedia article I linked in my Comment above gives more information.
Here is a simulation for $n = 10$ and $\theta = 5.$
th = 5; n = 10
th.mle = -n/replicate(10^6, sum(log(rbeta(n, th, 1))))
mean(th.mle)
## 5.555069 # aprx expectation of th.mle > th = 5.
median(th.mle)
## 5.172145
The histogram below shows the simulated distribution of $\hat \theta.$
The vertical red line is at the mean of that distribution, and the green
curve is its kernel density estimator (KDE). According to the KDE, its mode is near $4.62.$
den.inf = density(th.mle)
den.inf$x[den.inf$y==max(den.inf$y)]
## 4.624876
hist(th.mle, br=50, prob=T, col="skyblue2", main="")
abline(v = mean(th.mle), col="red")
lines(density(th.mle), lwd=2, col="darkgreen")
Addendum on Parametric Bootstrap Confidence Interval for $\theta:$
In order to find a confidence interval (CI) for $\theta$ based on MLE $\hat \theta,$ we would like to know the distribution of $V = \frac{\hat \theta}{\theta}.$ When that distribution is not
readily available, we can use a parametric bootstrap.
If we knew the distribution of $V,$ then we could find numbers $L$ and $U$ such that
$P(L \le V = \hat\theta/\theta \le U) = 0.95$ so that a 95% CI would be of the form
$\left(\frac{\hat \theta}{U},\, \frac{\hat\theta}{L}\right).$ Because we do not know the distribution of $V$ we use a bootstrap procedure to get serviceable approximations $L^*$ and $U^*$ of $L$ and $U.$ respectively.
To begin, suppose we have a random sample of size $n = 50$ from $\mathsf{Beta}(\theta, 1)$
where $\theta$ is unknown and its observed MLE is $\hat \theta = 6.511.$
Entering, the so-called 'bootstrap world'. we take repeated 're-samples` of size $n=50$
from $\mathsf{Beta}(\hat \theta =6.511, 0),$ Then we we find the bootstrap
estimate $\hat \theta^*$ from each re-sample. Temporarily using the observed
MLE $\hat \theta = 6.511$ as a proxy for the unknown $\theta,$ we find a large number $B$ of re-sampled values $V^* = \hat\theta^2/\hat \theta.$ Then we use quantiles .02 and .97 of
these $V^*$'s as $L^*$ and $U^*,$ respectively.
Returning to the 'real world'
the observed MLE $\hat \theta$ returns to its original role as an estimator, and the
95% parametric bootstrap CI is $\left(\frac{\hat\theta}{U^*},\, \frac{\hat\theta}{L^*}\right).$
The R code, in which re-sampled quantities are denoted by .re
instead of $*$, is shown below.
For this run with set.seed(213)
the 95% CI is $(4.94, 8.69).$ Other runs with unspecified
seeds using $B=10,000$ re-samples of size $n = 50$ will give very similar values. [In a real-life application, we would not know whether this CI covers the 'true' value of $\theta.$ However,
I generated the original 50 observations using parameter value $\theta = 6.5,$ so in this demonstration we
do know that the CI covers the true parameter value $\theta.$ We could have used the
probability-symmetric CI with quantiles .025 and .975, but the one shown is a little shorter.]
set.seed(213)
B = 10000; n = 50; th.mle.obs=6.511
v.re = th.mle.obs/replicate(B, -n/sum(log(rbeta(n,th.mle.obs,1))))
L.re = quantile(v.re, .02); U.re = quantile(v.re, .97)
c(th.mle.obs/U.re, th.mle.obs/L.re)
## 98% 3%
## 4.936096 8.691692
For this example
$$L(\theta;x_i)=\theta^{2n}\cdot \prod_{i=1}^n x_i\cdot e^{-\theta
\sum_{i=1}^nx_i}$$
This is not right. We have $f(x)=\theta^2 x e^{-\theta x}$ Now we calculate the product for every $x_i$
$$ L(\theta;x_i)=\prod_{i=1}^n \theta^2 x_i\cdot e^{-\theta x_i}=\theta^{2n}\cdot \prod_{i=1}^n x_i\cdot e^{-\theta x_i}$$
You see that there is as yet no sigma sign involved. There is either an sigma sign or a product sign.
At the next step, taking logarithm, there is a mistake. It is right that the $\theta^{2n}$ becomes the summand $2n\cdot \ln(\theta)$. Now we calculate
$\ln\left(\prod\limits_{i=1}^n x_i\cdot \large{e^{-\theta x_i}}\right)$
Firstly we use the logarithm rule $\log(a\cdot b)=\log(a)+\log(b)$ to eliminate the product sign.
$$= \sum_{i=1}^n \ln \left( x_i\cdot \large{e^{-\theta x_i}} \right)$$
We use the same rule again for a further simplification.
$$= \sum_{i=1}^n \ln \left( x_i \right) + \sum_{i=1}^n \ln\left( \large{e^{-\theta x_i}} \right)$$
$$= \sum_{i=1}^n \ln \left( x_i \right) -\theta \sum_{i=1}^n x_i$$
With the summnand $2n\cdot \ln (\theta)$ we have
$$\ln \left(L(\theta;x_i)\right)=2n\cdot \ln (\theta) +\sum_{i=1}^n \ln \left( x_i \right) -\theta \sum_{i=1}^n x_i$$
Now the derivative w.r.t. $\theta$ is
$$\frac{2n}{\theta}-\sum_{i=1}^n x_i=0$$
For the rest there are no logarithm rules required.
Best Answer
Hint to get you going: $X^\beta$ is distributed exponentially with parameter $\alpha$ [use the "transformation technique" for this].
Transformation technique: $Y=X^\beta\Rightarrow s(y)=x=y^{1/\beta}\Rightarrow \frac{ds(y)}{dy}=\frac 1\beta y^{\frac1\beta-1}$
$g(y)=\alpha\beta (y^{1/\beta})^{\beta-1}e^{-\alpha(y^{1/\beta})^\beta}\cdot \frac 1\beta y^{\frac1\beta-1}=\alpha e^{-\alpha y}$
Thus $\sum_{i=1}^n X_i^\beta$ is Gamma(n, $\alpha$). Then you would try to calculate the expectation of $\hat\alpha$ and wrt to the gamma pdf, and then unbias it by multiplying with a $c\in\mathbb R$. Let $Z=\sum_{i=1}^n X_i^\beta$.
$$\begin{split}E(\hat\alpha)&=\int_0^\infty\frac n{z}\cdot\frac{\alpha^n}{(n-1)!}z^{n-1}e^{-\alpha z}dz\\ &=\alpha\frac n{n-1}\int_0^\infty \frac{\alpha^{n-1}}{(n-2)!}z^{n-2}e^{-\alpha z}dz\\ &=\alpha \frac{n}{n-1}\end{split}$$
I get: $$c=\frac{n-1}{n}$$
$$E\left(c\hat\alpha - \alpha\right)=0$$