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Let $\kappa$ be an infinite cardinal. Then we have

$$\beth_{\kappa+1} = \prod_{\alpha<\kappa} \beth_{\alpha} = \beth_{\kappa}^{\kappa}$$

The only non-trivial inequality is the first less-than-or-equal-to which can be obtained by coding each $X \in \mathcal{P}(\beth_{\kappa}) \cong \beth_{\kappa+1}$ as the map $f \in \prod_{\alpha<\kappa} \beth_{\alpha}$ defined by $f(\alpha)=g_{\alpha}(X \cap \beth_{\alpha-1})$ for non-zero non-limit $\alpha$, where $g_{\alpha}: \mathcal{P}(\beth_{\alpha-1})\rightarrow \beth_{\alpha}$ is a fixed bijection, and $f(\alpha)=0$ for other $\alpha$.

I need to use this identity in a manuscript (intended for non-set theorists) and want to avoid including a full proof to save space. Does anyone know of any textbook which contains this as a fact? The closest one I was able to find is Exercise I.13.33 in Kunen's (new) book which only covers the case $\kappa=\omega$. (Worst case, I'll write down the proof myself but given Kunen's exercise, this product shouldn't be that unusual and must have been written down somewhere...)

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  • $\begingroup$ This does not seem correct to me. For many $\kappa$, we have $\beth_\kappa^\kappa=\beth_\kappa$. $\endgroup$ Sep 29, 2020 at 22:21
  • $\begingroup$ @AndrésE.Caicedo: The cofinality of $\beth_\kappa$ is the cofinality of $\kappa$. So that can't be right. $\endgroup$
    – Asaf Karagila
    Sep 29, 2020 at 22:23
  • $\begingroup$ @AndrésE.Caicedo: $\beth_{\kappa}^{\kappa} \leq \left(2^{\beth_{\kappa}}\right)^{\kappa} = 2^{\beth_{\kappa} \cdot \kappa}=2^{\beth_{\kappa}}=\beth_{\kappa+1} \leq \prod_{\alpha < \kappa}...$. So the result should hold if I am not making some obvious mistake with the inequality that I explained. $\endgroup$
    – Burak
    Sep 29, 2020 at 22:27
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    $\begingroup$ Now that I see my foolishness, this is really quite close to the manipulations involved in the combinatorial proofs of Silver's theorem (a quick argument as you suggest is in the chapter on COmbinatorics by Kunen in the Handbook of Mathematical Logic). Maybe look at the Galvin-Hajnal paper, or the book on the partition calculus by Erdős et al. $\endgroup$ Sep 29, 2020 at 22:50
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    $\begingroup$ By the way, "non-zero non-limit $\alpha$" is just "a successor $\alpha$", or even "$\alpha=\beta+1$". $\endgroup$
    – Asaf Karagila
    Sep 29, 2020 at 23:10

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