How to show that the Cohen forcing adding arbitrary many reals adds no dominating real

Question

Let $\lambda$ be any infinite cardinal and let $Fn(\lambda, 2)$ be the set of finite partial functions from $\lambda$ into $2$. This is a forcing notion adding $\lambda$ many Cohen reals. It is a well-known fact that this poset does not add a dominating real, but I have difficulty seeing it, or finding a proof.

Answer 1

First we prove it for adding one Cohen real. Let $\{ p_i : i < \omega \}$ enumerate the Cohen forcing. Let $\dot f$ be a name for a function in $^\omega \omega$. For $i < \omega$, choose $q_i \leq p_i$ such that $q_i$ decides $f(i) = j_i$. Define $g(i) = j_i$. Suppose $p \Vdash (\forall m > n) \dot f(m) > \check g(m)$. But let $i > n$ be such that $p_i \leq p$, so that $q_i \Vdash \check g(i) = \dot f(i)$, a contradiction.

Now any real $f$ added by $Fn(\lambda,2)$ is added by $Fn(X,2)$, where $X$ is countable. So there is some ground model real $g$ that $f$ does not dominate.