I know that if $X_1 \sim \operatorname{Geo}(p_1)$, and $X_2 \sim \operatorname{Geo}(p_2)$, then $\min{X_1,X_2} \sim \operatorname{Geo}(1-q_1 q_2)$ where $q_i=1-p_i$.
Does it holds for any number of geometric random variables? And in addition, they need to be independent?
Without computations: a $\mathcal{G}(p)$ random variable describes the time of the first success when you repeat independent experiments with success probability $p$. The minimum of independent $\mathcal{G}(p_1), \dots, \mathcal{G}(p_n)$ is the time of the first success of any of $n$ independent experiments with success probability $p_1, \dots, p_n$. Each time, there is a probability $1 - q_1 \cdots q_n$ that any of the $n$ experiments is a success (since there is a probability $q_1 \cdots q_n$ that they all are failures), so the time of the first success of any of the $n$ experiments is indeed geometric with parameter $1 - q_1 \cdots q_n$.
By independence, joint probability is a product of probabilities, so you need to a count for each of $n$ TV having a success in i$th$ trial and the remaining rvs having $i$ or more failures. This is where you use the independence property.
For $i=1,2,\dots,m$ and $n\in\mathbb N^+$ let $B_{i,n}$ be independent random variables having Bernoulli distribution with parameter $p_i$.
Then $X_i:=\min\{n\mid B_{i,n}=1\}$ has geometric distribution with parameter $p_i$.
For $X:=\min\{X_1,\dots,X_m\}$ we find:$$X=\min\{n\mid B_n=1\}\text{ where }B_n=1-\prod_{i=1}^m(1-B_{i,n})\tag1$$
Again the $B_n$ have Bernoulli distribution and this with parameter $1-\prod_{i=1}^m(1-p_i)$.
So $(1)$ makes clear that $X$ has geometric distribution with parameter $1-\prod_{i=1}^m(1-p_i)$.
