Suppose $M \subseteq N$ are models of ZFC such that $(ORD^\omega)^M = (ORD^\omega)^N$. Let $\langle P_n : n \in \omega \rangle$ be a sequence of countably closed partial orders in $M$, and let $\langle G_n : n \in \omega \rangle$ be a sequence of filters in $N$ such that for each $n$, $G_n$ is $P_n$generic over $M[G_0,...,G_{n1}]$. Is $\Pi_{n \in \omega} G_n$ generic for $\Pi_{n \in \omega} P_n$ over $M$?

This is a very nice problem, but unfortunately the answer can be negative. Let me describe a counterexample. Consider the forcing to add $\omega$ many Cohen subsets of $\omega_1$. So $P_n=\text{Add}(\omega_1,1)$ adds one Cohen subset to $\omega_1$ and the (full support) product $\Pi_n P_n$ is $\text{Add}(\omega_1,\omega)$. Suppose that $G\subset\Pi_n P_n$ is $M$generic for the product forcing, and consider the two models $M\subset N=M[G]$. Since the forcing is countably closed, it adds no new $\omega$sequences over $M$. We may think of $G$ as filling in a $\omega\times\omega_1$ matrix with $0$s and $1$s. Generically, there will be many allzero rows, that is, rows having zeros all the way across, so that $G(n,\alpha)=0$ for all $n$, where this is the $\alpha^{th}$ row. Let us define $G^\ast$ to be just like $G$ in every column, except that in any such allzero row in $G$, we change the first bit to a $1$ in the first column in $G^\ast$, leaving the rest of the row all $0$s. This operation ensures that $G^\ast$ has no allzero rows, and thus ensures that $G^\ast$ is definitely not $M$generic for the product forcing. But meanwhile, I claim that this operation does not affect the $M$genericity of any finite number of the columns of $G^\ast$. For this, it is an elementary exercise to see that for any dense set $D$ for the forcing in the first $n$ columns, there is a dense set $E$ in the full product, such that the operation applied to conditions in $E$ gives a condition in $D$. So generically, $G$ is such that $G^\ast$ will have its first $n$ factors in $D$. Thus, every finitely many factors of $G^\ast$ are $M$generic, but the whole product $G^\ast$ is not $M$generic; so it is a counterexample. 

