"Forcing a Cohen real inside a Cohen real" not a product forcing My professor mentioned that binary iterated forcings $P *\dot{Q}$ cannot always be written as product forcings $P \times R$ for some forcing notion $R$, hence iterated forcing is a proper generalization.
I see that an easy counterexample is $P * \dot{Q}$ where $P$ is the Cohen forcing $\dot{Q}$ is the name of the Sacks forcing notion in the extension; by minimality of Sacks it is minimal over another extension, but if it were $P \times R$ then by the Product Lemma it would not be (since Cohen forcing is not minimal).
However, the example my professor gave was that if we do the Cohen forcing, and then "force a Cohen real inside of one of the new Cohen reals" (force a Cohen subset of $\{n : s(n) = 1 \}$ where $s$ is a Cohen real), then that is not the same as $P\times R$ for any forcing notion $R$ where $P$ is the Cohen forcing.
Why is this?
 A: Good question!
Let me begin with apparently claiming that your professor is wrong: the "Cohen-in-Cohen" extension your professor describes is exactly the product-of-two-Cohens extension.
Let $G$ be Cohen-generic over $V$, let $\mathbb{P}$ be Cohen forcing, and let $\mathbb{Q}$ be Cohen-in-$G$ forcing. Let $H$ be $\mathbb{Q}$-generic over $V[G]$, and let $$J=\{i: \mbox{the $i$th element of $G$ is in $H$}\}.$$ 
Note that the Cohen poset is the same across all forcing extensions - this is not generally true. For example, let $\mathbb{S}_0$ be Sacks forcing in $V$, $G$ be $\mathbb{S}_0$-generic, and $\mathbb{S}_1$ be Sacks forcing in $V[G]$. Then $V[G]\models\mathbb{S}_0\not\cong\mathbb{S}_1$. This is why the square Sacks extension is different from the Sacks-over-Sacks extension.
I claim that $J$ is Cohen-generic over $V[G]$. To see that $J$ is Cohen over $V[G]$, note that there is an obvious isomorphism from $\mathbb{Q}$ to $\mathbb{P}$, and that this isomorphism lives in $V[G]$ and carries $H$ to $J$.
So what? Well, we have $V[G][H]=V[G][J]$ (since we can recover $H$ from $G$ and $J$, and vice-versa). So $\mathbb{P}*\dot{\mathbb{Q}}$ and $\mathbb{P}^2$ yield the same generic extensions. So I would say "$\mathbb{P}^2$ and $\mathbb{P}*\dot{\mathbb{Q}}$ are the same"
. . . in a sense. The issue is what "same" means, here. The two posets $\mathbb{P}^2$ and $\mathbb{P}*\dot{\mathbb{Q}}$ are not, in fact, isomorphic; so it depends what notion of "sameness" we're looking for here. The example you give is a better one, since we can distinguish the two posets even at the level of generic extensions they yield.
