User blog comment:SuperJedi224/Question/@comment-5029411-20141124132603/@comment-11227630-20141126050248

It's $$\theta(\Omega^\omega,1)$$, because it's also the supremum of $$\theta(\alpha,\theta(\Omega^\omega,0)+1)$$ for all $$\alpha<\theta(\Omega^\omega,0)$$.