Minimum of $n$ geometric random variables

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 $1 - q_1 cdots q_n$.

For $i=1,2,dots,m$ and $ninmathbb N^+$ let $B_{i,n}$ be independent random variables having Bernoulli distribution with parameter $p_i$.

Then $X_i:=min{nmid B_{i,n}=1}$ has geometric distribution with parameter $p_i$.

For $X:=min{X_1,dots,X_m}$ we find:$$X=min{nmid B_n=1}text{ where }B_n=1-prod_{i=1}^m(1-B_{i,n})tag1$$

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 $1-prod_{i=1}^m(1-p_i)$.

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.