If I have a statement that says,
$I\implies J\implies K$.
Can I say that,
since, $J \implies K$ and $J$ is true, $K$ is true as well.
Or Do I have to prove that $I$ is true as well to say so?
Sometimes we briskly express the three metalogical claims that (1) implies (2), and (2) implies (3), and (3) implies (1) in the compressed form $$(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$$ So in this case, to be sure, if e.g. you already have an independent proof that (2) is true, you could extract the claim that $(2) \Rightarrow (3)$ and conclude that (3).
However we never make that contraction inside the propositional calculus: i.e. we never contract $$p \to q \land q \to r$$ to $$p \to q \to r.$$ Depending on your official rules for dropping brackets, the latter is either illegitimate or is to be read as $$p \to (q \to r)$$ which means something quite different. And to conclude $r$ given the latter, you'd need to invoke both the truth of $p$ and the truth of $q$.
$J\Rightarrow K$ means that when $J$ is true, $K$ must be true as well, so, yes, this would be a reasonable deduction given $J$.