Is the following consistent?


There are definable class forcing notions $\lbrace \mathbb{P}\_{n}\rbrace\_{n\in \omega}$ such that:


1. The product of any finitly many of them preserves $\text{ZFC}$ and all cardinals.
2. The $\prod\_{n\in \omega}\mathbb{P}\_{n}$ collapses all uncontable cardinals to $\omega$.



Countable Product of Class Forcing Notions
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](http://jdh.hamkins.org/a-forcing-class-and-its-modal-logic/), 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 $\omega$. 


