Distribution of transformation Suppose $X_1,\ldots,X_n$ are i.i.d. $\mathcal U(0,1)$. I am looking for the asymptotic distribution of $$T_n = \prod_{i=1}^n [e{X_i}]^{1/\sqrt{n}} \>.$$
 A: Since the OP never answered his (or her) own question:
Following whuber's sensible advice, we intend to consider the logarithm of the thing. First we note that since $X_i \sim U(0,1)$ we have that
$$Z = -\ln X_i \Rightarrow f_z(z) = e^{-z}$$
i.e. $Z$ follows an Exponential distribution with $E(Z) = 1, \text{Var}(Z) = 1$. So the sample mean $\bar Z_n$ from an i.i.d. sample has $E(\bar Z_n) = 1, \text{Var}(\bar Z_n) = 1/n$.  Keeping these in mind we plunge into
$$\ln T_n = \ln\left[\prod_{i=1}^n [e{X_i}]^{1/\sqrt{n}} \right] = \ln\left[(e^{1/\sqrt{n}})^n \prod_{i=1}^n {X_i}^{1/\sqrt{n}} \right]$$
$$=\sqrt n\ln e + \frac1{\sqrt n}\sum_{i=1}^n\ln X_i = \sqrt n\left(1-\frac 1n\sum_{i=1}^n\left(-\ln X_i\right)\right) $$
$$\Rightarrow \ln T_n = \sqrt n\left(1-\frac 1n\sum_{i=1}^nZ_i\right) = -\frac{\bar Z_n-E(\bar Z_n)}{SD(\bar Z_n)} $$
The final expression is the subject matter of the classical Central Limit Theorem (the minus sign does not bother us). Since $X$'s are i.i.d. so are their functions, and then
$$\ln T_n \xrightarrow{d} T^* \sim N(0,1)$$
Using the Continuous Mapping (Mann-Wald) theorem $Y_n \xrightarrow{d} Y \Rightarrow h(Y_n) \xrightarrow{d} h(Y)$ for $h$ continuous
and setting $h(y) = e^y$ we obtain
$$e^{\ln T_n} = T_n \xrightarrow{d} e^{T^*}$$
Since $T^*$ is a standard normal random variable, $e^{T^*}$ is a log-normal RV, so
$$T_n \xrightarrow{d} LN(0,1),\;\; E(T^*) = e^{1/2}, \text{Var}(T^*) = e^2 -e $$
