Countable Product of Class Forcing Notions Is the following consistent?
There are definable class forcing notions $\lbrace \mathbb{P}_{n}\rbrace_{n\in \omega}$ such that:


*

*The product of any finitly many of them preserves $\text{ZFC}$ and all cardinals.

*The $\prod_{n\in \omega}\mathbb{P}_{n}$ collapses all uncontable cardinals to $\omega$. 
 A: Yes, this can happen. 
Start with $V=L$ (but $V=HOD+GCH$ suffices), and assume there are no inaccessible cardinals. For each regular cardinal $\delta$, partition the ordinals of cofinality $\delta$ below $\delta^+$ into $\omega$ many disjoint stationary sets $\text{Cof}_\delta\cap\delta^+=\bigsqcup_n S^\delta_n$. Let $\mathbb{P}_n$ be the Easton support product that at stage $\delta$ shoots a club disjoint from $S^\delta_n$, destroying its stationarity, using the club-shooting forcing of $L$, which consists of closed bounded subsets disjoint from that set, ordered by end-extension. At each stage, this is $\lt\delta$-closed and $\leq\delta$-distributive, a result due to Baumgartner, Harrington and Kleinberg (the case of $\delta^+=\omega_1$ is explained in lemma 17 of this paper, but the general argument is similar). 
Thus, the forcing is gradually killing off components of the stationary partition. The usual Easton arguments, combined with the analysis of club-shooting forcing, show that each $\mathbb{P}_n$ preserves ZFC and all cardinals and cofinalities, and this is also true of any finite product of the $\mathbb{P}_n$'s. 
But the combined forcing of all $\mathbb{P}_n$ will kill off the entire partition, which collapses the cardinal. In particular, the full product will collapse every successor cardinal $\delta^+$, since the intersection of all the clubs at any given stage must be empty. It follows that every cardinal will be collapsed to $\omega$. 
