I'm suppose to find the UMVUE for the variance $\theta^2$ of the exponential distribution $Exp(\theta) \sim f(x,\theta)=\frac{1}{\theta}\exp(-x/\theta)$
The hint is to use $\bar{X}^2$.
In the brief solution, it expressed $E(\bar{X}^2)=\frac{n+1}{n}\theta^2$
Then it stated that after bias correction, by lehmann-scheffe: $\frac{n(\bar{X}^2)}{n+1}=\frac{T^2}{n(n+1)} = $ UMVUE of variance($\theta^2$)
My problem is, I don't follow what bias correction step it applied before Lehmann-Scheffe? What did it use as the initial unbiased estimator to apply L-S Theorem? How did it turn $E(\bar{X}^2)$ to $\bar{X}^2$? How L-S Theorem would be needed to have $\bar{X}^2 = T^2/n^2$
The hint is correct but you can use also this one that, in my humble opinion, looks more suitable (the more basic is the estimator the better it is):
$$T=\left(\Sigma_i X_i\right)^2$$
This because $Y=\Sigma_i X_i$ is complete and sufficient (CSS), thus the goal now is to construct an unbiased estimator based on the CSS.
$$\mathbb{E}[T]=\int_0^{\infty} y^2\frac{1}{\theta^n \Gamma(n)}y^{n-1}e^{y/\theta}dy=$$
$$=\frac{\Gamma(n+2)\theta^2}{\Gamma(n)}\underbrace{\int_0^{\infty} y^2\frac{1}{\theta^{n+2} \Gamma(n+2)}y^{(n+2)-1}e^{y/\theta}dy}_{=1}=(n+1)n\theta^2$$
thus the UMVUE is
$$T_{\text{UMVUE}}=\frac{\left(\Sigma_i X_i\right)^2}{(n+1)n}$$
It is self evident that this estimator matches with the one stated in you textbook
$$\frac{n(\overline{X})^2}{n+1}$$