Given $$A \wedge B \to C \equiv A \to B \to C$$
We want to show that $$\{ A \to B \to C\} \vdash A \wedge B \to C$$ by constructing a derivation using the natural deduction system $\mathcal{N}_{PL}$.
My question: How do i write down such a derivation? So far I just came up with a "prove" by transformation: \begin{align} &A \to (B \to C)\\ &A \to (\neg B \vee C)\\ &\neg A \vee \neg B \vee C\\ &(\neg A \vee \neg B) \vee C\\ &\neg (\neg A \vee \neg B) \to C\\ &A \wedge B \to C \end{align}
First, your expression $A\rightarrow B\rightarrow C$ is not a well formed formula, you need to bracket it somehow and it makes a difference - the material conditional isn't associative.
Now, when they say to prove it using natural deduction, they mean to prove it using only a specific set of rules, namely the rules of natural deduction.
Read these https://www.cs.cornell.edu/courses/cs3110/2013sp/lectures/lec15-logic-contd/lec15.html http://logicmanual.philosophy.ox.ac.uk/jsslides/ll6.pdf