Let $X,Y$ be independent from the parameter $0 < p < 1$ distributed random variable on a discrete probability space $(\Omega, \mathcal{F}, \mathbb P)$. Show that $\min\{X,Y\}$ is geometrically distributed to $1-(1-p)^{2}$.
I am having trouble understanding what it is my end result has to be, in order to show that it is indeed geometrically distributed. I mean for a geometric distribution $\mathbb P(k) = p(1-p)^{k-1}$ for parameter $p$, but how exactly am I showing a geometric distribution?
$\mathbb P( \min\{X,Y\} > k)=\mathbb P((X>k) \cap (Y>k))$ and then what?
In terms of the parameter, I've got: $(1-(1-p)^{2})(1-(1-(1-p)^{2}))=p^{2}(2-p)^{2}$
I do not understand how the two work together...
\begin{align} P(\min(X,Y) > k) &= P(X > k, Y>k) \\ &= P(X > k)P(Y>k) \\ &= P(X >k)^2 \\ &=((1-p)^2)^{k} \end{align} Hence the CDF is $1-((1-p)^2)^k=1-(1-\color{blue}{(1-(1-p)^2)})^k$
Hence it is a geometric distribution (we just have to compare with the CDF of a geometric distribution) with success probability $1-(1-p)^2.$