An excercise of my book asks me to prove that $P \vdash \sim \sim P$ by means of Łukasiewicz's axiom system. I've been able to show that $\sim\sim P \vdash P$, yet this doesn't seem to be of any help.
The axioms, as defined in my book (Logic for Philosophers), are all substitution instances of: $$(p\rightarrow(q\rightarrow p))$$ $$(p\rightarrow(q\rightarrow r))\rightarrow((p\rightarrow q)\rightarrow (p\rightarrow r))$$ $$(\sim p \rightarrow \sim q)\rightarrow ((\sim p \rightarrow q)\rightarrow p)$$
The only rule of inference is Modus Ponens.
Hmm, the best known Łukasiewicz's axiom system has $(\neg q \to \neg p) \to (p \to q)$ as an axiom in place of your third.
Fortunately, you can easily show $\vdash (\neg q \to \neg p) \to (p \to q)$ using your $(\neg p \to \neg q)\to((\neg p \to q)\to p)$:
Start with $\neg p \to \neg q$ and $q$. By axiom 1, you have $q \to (\neg p \to q)$, and so by MP: $\neg p \to q$, and now it is two MP's with your Axiom 3 to get $p$.
So, we have $\neg p \to \neg q, q \vdash p$, and thus by the Deduction Theorem $\neg p \to \neg q \vdash q \to p$
With that, and with your already shown $\neg \neg p \vdash p$ it's trivial, as by the Deduction Theorem you get $\vdash \neg \neg \neg p \to \neg p$, and with this axiom 3 you have $(\neg \neg \neg p \to \neg p) \to (p \to \neg \neg p)$, so a simple MP gives you $p \to \neg \neg p$