User blog comment:Vel!/A medley of countability proofs/@comment-5029411-20130516233018/@comment-1605058-20130517164841

Say we have found bijection \(f_\alpha :\mathbb{N}\rightarrow\alpha\). We wish to find similar bijection \(f_{\omega^\alpha}\). Define \(f_{\omega^\alpha}(p_n):=\omega^{f_\alpha(n)}\) (here we either have to take p_0=2, p_1=3 etc.) and \(f_{\omega^\alpha}(a\cdot b):=f_{\omega^\alpha}(a)+f_{\omega^\alpha}(b)}. It can be proven (via FT of arithmetic) that this function is injective and surjective, thus bijective. If we take \(f_\omega(n)=n\) we get precisely FB100Z's construction. Given FB's \(f_\omega_2\) we get what Aarex asked for, and we can inductively repeat construction up to \(\varepsilon_0\).