Exercise 15.15 of Jech: forcing GCH Jech uses the following forcing notion to force GCH. For each $\alpha$, let $P_\alpha$ be the notion of forcing that collapses $\beth_{\alpha+1}$ to $\beth_\alpha^+$. $P$ is then the Easton product of these notions. For simplicity, we'll assume there are no inaccessible cardinals. Let $P^{< \alpha}$ be the product below $\alpha$ and let $P^{\leq\alpha}$ be the product up to and including $\alpha$. So, $P^{<\alpha}$ is just the set of arbitrary functions from $\alpha$ to the corresponding partial orders $P_\beta$ for $\beta<\alpha$.
In part, my strategy is to prove that $P^{\leq \alpha}$ forces CH at $\beth_\beta$, for $\beta\leq\alpha$.
Let $\lambda$ be a limit ordinal and assume I've shown this for $\alpha<\lambda$ and I want to show it for $P^{\leq \lambda}$. I know how to show that $P^{<\lambda}$ forces CH at $\beth_\alpha$, for $\alpha<\lambda$. So the obvious thought is to force with $P_\lambda$ first, since that is $\beth_\lambda$-closed and will preserve all the relevant properties of $P^{<\lambda}$ whilst making $\beth_{\lambda+1} = \beth_\lambda^+$. So, all I need to show is that when I then force with $P^{<\lambda}$, $\beth_{\lambda+1} = \beth_\lambda^+$ is preserved. That's what I'm having difficultly seeing. In particular, I can't see why $\beth_{\lambda+1}$ might not change.
 A: The solution to this exercise is quite long. It can be broken down to the following steps.
Let $P_\alpha$ be $\text{Col}(\beth_\alpha^+, \beth_{\alpha+1})$. As in the hint, let $P$ be the Easton class product of the $P_\alpha$'s, i.e. bounded support below inaccessibles. And $P^{>\alpha}, P^{\geq\alpha}, P^{\leq\alpha}$ are defined in the familiar way. Also notice that Jech defines $\kappa$-closed with quantification over sequences of length $\leq\kappa$. Kunen's book would call this $\kappa^+$-closed. I am using Jech's convention for consistency with the lemmas he refers to. Another notational convention: in this answer the cardinal terms without superscript are evaluated in $V$.
Lemma 1. For each $\alpha$, $P_\alpha$ is $\beth_\alpha$-closed and has the $\beth_{\alpha+1}^+$-cc (the latter holds because by cardinal arithmetic calculation, $\beth_{\alpha+1}^{<\beth_\alpha^+}=\beth_{\alpha+1}$ and $|P_\alpha|=\beth_{\alpha+1}$).
Lemma 2. For each $\alpha$, $P^{\geq\alpha}$ is $\beth_\alpha$-closed. So in particular $P^{>\alpha}$ is $\beth_{\alpha+1}$-closed. Also, the whole product $P$, viewed as $P^{\geq 0}$ is then $\omega$-closed.
Lemma 3. For each $\alpha$, $P^{\leq\alpha}$ has the $\beth_{\alpha+1}^+$-cc.
1-3 are hints given in the exercise, the following are not.
Lemma 4. Each $(\beth_\alpha^+)^V$ is a cardinal in $V[G]$.
Proof. We prove the lemma by considering different cases. The case of $\alpha=0$ follows from $P$ being $\omega$-closed. For the case of $\alpha+1$, suppose for contradiction that $\beth_{\alpha+1}^+$ is not a cardinal in $V[G]$. Then in $V[G]$ we fix a bijection $f:\beth_{\alpha+1}\leftrightarrow \beth_{\alpha+1}^+$. By general facts of product forcing, we may view $V[G]$ as $V[G^{\leq\alpha}\times G^{>\alpha}]$, so Lemma 1 and 2 mean that we can apply Lemma 15.19 in Jech and deduce $f\in V[G^{\leq\alpha}]$. This means $\beth_{\alpha+1}^+$ is already not a cardinal in $V[G^{\leq\alpha}]$. This contradicts the fact that $P^{\leq\alpha}$ has the $\beth_{\alpha+1}^+$-cc. Almost the same argument, viewing $V[G]=V[G^{<\alpha}\times G^{\geq\alpha}]$ takes care of the case when $\alpha$ is inaccessible.
So it remains to show the case of $\alpha$ a singular limit. This means that $\beth_\alpha$ is singular in both $V$ and $V[G]$. Now suppose for contradiction that $\beth_\alpha^+$ is not a cardinal in $V[G]$. Consider what $V[G]$ thinks is the cardinality of $\beth_\alpha^+$. In $V[G]$, we have $|\beth_\alpha^+|^{V[G]}<\beth_\alpha^+$. Being a cardinal is downward absolute, so $|\beth_\alpha^+|^{V[G]}$ is a cardinal in $V$ as well. But since there's no cardinal between $\beth_\alpha$ and $\beth_\alpha^+$, we see that $|\beth_\alpha^+|^{V[G]}$ is at most $\beth_\alpha$. Notice that in $V[G]$, $|\beth_\alpha^+|^{V[G]}$ is at least $\beth_\alpha$, because $\beth_\alpha$ injects into $\beth_\alpha^+$, which in turn injects into $|\beth_\alpha^+|^{V[G]}$. This means $\text{cf}^{V[G]}(\beth_\alpha^+)\leq\text{cf}^{V[G]}(\beth_\alpha)<\beth_\alpha$. This tells us that $V[G]$ thinks $\beth_\alpha^+$ is singular. Splitting $V[G]$ into $V[G^{\leq\alpha}\times G^{>\alpha}]$ again, we can show using Lemma 15.19 that it must already be singular in $V[G^{\leq\alpha}]$, which contradicts the the cc property of $P^{\leq\alpha}$. $\square$
Lemma 5. In $V[G]$, each $(\beth_\alpha)^V$ has cardinality $\aleph_\alpha^{V[G]}$.
Proof. We show this by induction on $\alpha$. The case $\alpha=0$ is by absoluteness of $\omega$. The limit case is straightforward. So let's assume the lemma holds for $\alpha$, and consider $\alpha+1$.
First, notice that by the inductive assumption, we have that
$$
\aleph_\alpha^{V[G]}=|\beth_\alpha|^{V[G]}\leq \beth_\alpha<\beth_\alpha^+=|\beth_\alpha^+|^{V[G]}.
$$
where the last equality is by Lemma 4. Also, every $V[G]$-cardinal $\kappa<\beth_\alpha^+$ is a $V$-cardinal by absoluteness, so $\kappa\leq|\beth_\alpha|^{V[G]}=\aleph_\alpha^{V[G]}$. This shows $\beth_\alpha^+=\aleph_{\alpha+1}^{V[G]}$.
Now consider the projection $G_\alpha$ of $G$ onto $P_\alpha$. This gives us in $V[G]$ a function $f=\bigcup G_\alpha$ from $\beth_\alpha^+$ onto $\beth_{\alpha+1}$. THat is, $|\beth_\alpha^+|^{V[G]}=|\beth_{\alpha+1}|^{V[G]}$.  Putting things together, we've shown $|\beth_{\alpha+1}|^{V[G]}=\aleph_{\alpha+1}^V{[G]}$. $\square$
We are now ready to show the main claim.

Claim. $V[G]$ satisfies the GCH.

Proof. It suffices to show that $(2^{{\aleph_\alpha}^{V[G]}})^{V[G]}\leq \aleph_{\alpha+1}^{V[G]}$. We consider three different cases: successor, inaccessible, singular limit.
The successor case is a straightforward cardinality calculation: $(2^{{\aleph_{\alpha+1}^{V[G]}}})^{V[G]}=(2^{\beth_{\alpha+1}})^{V[G]}$ by Lemma 5. And $(2^{\beth_{\alpha+1}})^{V[G]}=(2^{\beth_{\alpha+1}})^{V[G^{\leq\alpha}]}$ by Lemma 15.19. Furthermore, $(2^{\beth_{\alpha+1}})^{V[G^{\leq\alpha}]}$ is at most $|B(P^{\leq\alpha})|^{\beth_{\alpha+1}}$ by counting names. Since each element of $B(P^{\leq\alpha})$ can be determined by an antichain from $P^{\leq\alpha}$, we have:
$$
|B(P^{\leq\alpha})|^{\beth_{\alpha+1}}\leq|P^{\leq\alpha}|^{\beth_{\alpha+1}}=\beth_{\alpha+1}^{\beth_{\alpha+1}}=2^{\beth_{\alpha+1}}=\beth_{\alpha+2}=\aleph_{\alpha+2}^{V[G]}
$$
This shows that $(2^{{\aleph_{\alpha+1}^{V[G]}}})^{V[G]}\leq \aleph_{\alpha+2}^{V[G]}$.
The inaccessible case is similar, by breaking $V[G]$ instead into $V[G^{<\alpha}\times G^{\geq\alpha}]$.
Now suppose $\alpha$ is a singular limit ordinal. Then $\beth_\alpha$ is singular. Take a confinal sequence $(\alpha_\beta\mid \beta<cf(\alpha)=\gamma)$ supping to $\alpha$ and consider $\beth_\alpha$ as $\bigcup_{\beta}\beth_{\alpha_\beta}$. We notice that $(2^{\beth_\alpha})^{V[G]}$ is at most $|\beth_\alpha|^\gamma$: if $X\in \mathcal{P}^{V[G]}(\beth_\alpha)$, then we map $X$ to the "stratifiying" sequence $(X\cap \beth_{\alpha_\beta}\mid \beta<\gamma)$. This map is injective. Taking stock, we've shown that $V[G]$ thinks $2^{|\beth_\alpha|}$ is at most $|\beth_\alpha|^\gamma$ (the latter being evaluated in $V[G]$). That is,
$V[G]\models 2^{{|\beth_{\alpha}|}}\leq\prod_{{\beta<\gamma}} 2^{{|\beth_{{\alpha_{\beta}}}|}}\leq\prod_{{\beta<\gamma}}\aleph_{{\alpha_{\beta}+1}}=\left(\sup_{{\beta<\gamma}}\aleph_{{\alpha_{\beta}}}\right) ^{\gamma}={|\beth_{{\alpha}}|}^{\gamma}$
This latter cardinal $|\beth_\alpha|^\gamma$ is at most $|\beth_\alpha^{\beth{\gamma+1}}|^{V[G]}$, which is in turn at most $|\beth_\alpha^{\beth{\gamma+1}}|^{V[G^{\leq\gamma}]}$. But now
$$
|\beth_\alpha^{\beth{\gamma+1}}|^{V[G^{\leq\gamma}]}\leq  |2^{\beth_\alpha\times\beth_{\gamma+1}}|^{V[G^{\leq\gamma}]} \leq |B(P^{\leq\gamma})^{\beth_\alpha\times\beth_{\gamma+1}}|= \beth_{\gamma+2}^{\beth_\alpha\times\beth_{\gamma+1}}=\beth_{\alpha+1}
$$
Using Lemma 5 again, $|\beth_{\alpha+1}|^{V[G]}=\aleph_{\alpha+1}^{V[G]}$, we conclude that $(2^{{\aleph_{\alpha}^{V[G]}}})^{V[G]}\leq{\aleph_{\alpha+1}}^{V[G]}$ for singular limit $\alpha$. This finishes the proof.
