Let $U_{i}, i=1,2, \ldots$ be independent uniform random variables in $(0,1)$. Also $X$ is a discrete random variable whose pdf is given by: $$P(X=x)=\frac{1}{(e-1)x !}, x=1,2,3, \ldots$$ Find the CDF of $Z$ defined as: $Z=\min \left\{U_{1}, \ldots, U_{X}\right\}$
My try:
We have $F_Z(z)=P(Z\leq z)=1-P(Z>z)=1-P(U_1>z,U_2>z,....U_X>z)=1-(P(U_1>z))^X$
So we get $$F_Z(z)=1-(1-P(U_1\leq z))^X=1-(1-z)^X$$ But i am stuck here, how to use PMF of $X$?
The computation becomes a bit clearer if we write the hierarchical model explicitly. Let $$Z \mid X = \min(U_1, \ldots, U_X)$$ represent the conditional random variable of the first order statistic given $X$ observations. Then as you already noted, the conditional cumulative distribution of $Z$ given $X$ is $$\Pr[Z \le z \mid X] = \begin{cases} 0, & z < 0 \\ 1 - (1-z)^X, & 0 \le z \le 1 \\ 1, & z > 1. \end{cases}$$
Then by the law of total probability, the unconditional or marginal distribution of $Z$ is $$\Pr[Z \le z] = \sum_{x=1}^\infty \Pr[Z \le z \mid X = x]\Pr[X = x].$$
This is the piece of information you were missing. We then have
$$\Pr[Z \le z] = \sum_{x=1}^\infty \frac{1 - (1-z)^x}{(e-1)x!} = \frac{1}{e-1} \left( \sum_{x=1}^\infty \frac{1}{x!} - \sum_{x=1}^\infty \frac{(1-z)^x}{x!}\right).$$ Now using the fact that $$e^t = \sum_{x=0}^\infty \frac{t^x}{x!},$$ it is straightforward to finish the computation, being careful to note that lower index of summation of the desired sum begins at $1$, not $0$.