$\beth_\xi=\sum_{\eta<\xi}2^{\beth_{\eta}}$

Question

If $\xi>0$ is an ordinal, then $$\beth_\xi=\sum_{\eta<\xi}2^{\beth_{\eta}}.$$

This is my attempt:

By definition of the beth function, $$\beth_\xi=\sum_{\eta<\xi}\beth_\eta.$$

Therefore, since $2^{\beth_\alpha}=2^{\beth_{\alpha+1}}$, $$\beth_\xi=\sum_{\eta<\xi}\beth_\eta=\sum_{\eta<\xi}2^{\beth_{\eta-1}}=\sum_{\eta<\xi}2^{\beth_\eta}.$$

I think I can't do the first step, because $\xi$ must be a limit ordinal. Any hint, please? :(

Answer 1

Hint: $$ \sum_{\eta<\xi+1} \beth_\eta = \sum_{\eta\leq\xi} \beth_\eta = \beth_\xi $$