How do I show that $\vdash (\lnot \alpha \to \alpha) \to \alpha$?

Question

I want to show that: $\vdash (\lnot \alpha \to \alpha) \to \alpha$, and here's what I've tried:

1. Using the deduction theorem, this is the same as proving $\{(\lnot \alpha \to \alpha)\} \vdash \alpha$
2. $\alpha \to (\lnot \alpha \to \alpha)$ (axiom)
3. $(\alpha \to (\lnot \alpha \to \alpha)) \to ((\alpha \to \lnot \alpha) \to (\alpha \to \alpha)) $ (axiom)
4. $(\alpha \to \lnot \alpha) \to (\alpha \to \alpha)$ (Modus-Ponens on 2 and 3)

How do I go ahead from here? Thanks!

List of Axioms:

• $\alpha \to (\beta \to \alpha)$ (1)
• $(\alpha \to (\beta \to \gamma)) \to ((\alpha \to \beta) \to (\alpha \to \gamma))$ (2)
• $(\lnot \beta \to \lnot \alpha) \to (\alpha \to \beta)$ (3)

and Modus-Ponens is the sole rule of inference.