In addition to the two very nice formal answers, let me give an informal proof that $\neg\neg\neg A \Rightarrow \neg A$.
Recall that, by definition, $\neg B$ is an abbreviation for $B \Rightarrow \bot$.
Assume $\neg\neg\neg A$. We are to show $\neg A$, that is, $A \Rightarrow \bot$.
So assume $A$, we are to show $\bot$.
Since $A$, we have $\neg\neg A$. But we also have $\neg(\neg\neg A)$. Hence $\bot$.
This proof uses the lemma that $A \Rightarrow \neg\neg A$. This lemma can also be proven informally:
Assume $A$. We are to show $\neg\neg A$, that is, $(\neg A) \Rightarrow \bot$.
So assume $\neg A$. We are to show $\bot$.
Since $\neg A$ (which means $A \Rightarrow \bot$) and since $A$, we have $\bot$.
By the way, at no point did we use the principle of explosion (ex falso quodlibet). Hence our proofs are even valid in minimal logic, where $\bot$ doesn't have any special rules attached to it.