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Take the steps $Q_\alpha$ to be Cohen forcing, and begin by taking a generic filter $G$ for the usual finite-support product. So $G$ codes an $\omega$-sequence of Cohen reals $x_n\in 2^\omega$. Fix some $z\in 2^\omega$ coding an ordinal larger than the height of $M$. Define $y_n\in 2^\omega$ to be exactly the same as $x_n$ except that $y_n(0)=z(n)$. Since only one entry in $x_n$ has been changed, it is clear that $y_n$ is Cohen-generic over $M$ and that, for each natural number $n$, $\langle y_k:k<n\rangle$ is generic for $P_0*\dots*P_{n-1}$; let $G_n$ be the corresponding generic filter in $P_0*\dots*P_{n-1}$. The sequence $\langle G_n:n\in\omega\rangle$ is not in any extension of $M$ with the same ordinals, because it encodes $z$.
Take the steps $Q_\alpha$ to be Cohen forcing, and begin by taking a generic filter $G$ for the usual finite-support product. So $G$ codes an $\omega$-sequence of Cohen reals $x_n\in 2^\omega$. Fix some$z\in 2^\omega$coding an ordinal larger than the height of $M$. Define$y_n\in 2^\omega$ to be exactly the same as $x_n$ except that $y_n(0)=z(n)$. Since only one entry in $x_n$ has been changed, it is clear that $y_n$ is Cohen-generic over $M$ and that, for each natural number $n$, $\langle y_k:k<n\rangle$ is generic for $P_0*\dots*P_{n-1}$; let $G_n$ be the corresponding generic filter in $P_0*\dots*P_{n-1}$. The sequence $\langle G_n:n\in\omega\rangle$ is not in any extension of $M$ with the same ordinals, because it encodes $z$.