This result is obtained with Rahman
Theorem. Suppose there is an $\aleph_1$-complete ultrafilter $U$ over $\lambda.$ Then forcing with full product $\mathbb{P}=\prod_{i<\lambda}Add(\omega, 1)$ collapses $2^\lambda$ into $\aleph_0.$
Proof. Let $(r_i: i<\lambda)$ be the generic reals added by $\mathbb{P},$ and for each limit ordinal $\alpha,$ consider the reals $(r_{\alpha+n}: n<\omega)$ as an $\omega\times \omega$ matrix of $0, 1$'s as in Hamkins answer, and so as in Hamkins we can find for each $n<\omega$ a set $A_{\alpha, n} \subseteq [\alpha, \alpha+\omega).$ Let $A_n=\bigcup \{A_{\alpha, n}:\alpha$ is a limit ordinal $<\lambda \}.$
Working in $V$, define an equivalence relation $\sim$ on $P(\lambda)$ by $A\sim B$ iff $\{ \alpha<\lambda: lim(\alpha), A\cap [\alpha, \alpha+\omega)= B\cap [\alpha, \alpha+\omega)\} \in U.$
Claim. There are $2^\lambda$ many equivalence classes.
Now we show that for any $A \subseteq \lambda,$ there is $n_0<\omega$ such that $A \sim A_{n_0}.$
Given condition $p=(p_\alpha: \alpha<\lambda)\in \mathbb{P}$, there is $n_0 <\omega$ such that $X=\{\alpha< \lambda: dom(p_\alpha)=n_0\}\in U$ (use the $\aleph_1$-completeness of $U$). Now as in Hamkins argument, it is easily seen that we can extend $p$ to some condition $q$ forcing for all $lim(\alpha)\in X, A\cap [\alpha, \alpha+\omega)=A_{\alpha, n_0}$
The result follows immediately.
Corollary. Suppose $\lambda$ is a measurable cardinal, or is above the least strongly compact cardinal. Then forcing with full product $\mathbb{P}=\prod_{i<\lambda}Add(\omega, 1)$ collapses $2^\lambda$ into $\aleph_0.$