Iterated Forcing, to force $2^{\omega}=\kappa$ and $2^{\omega _1}=\lambda$ Hellow  i'm stuck on some details in this iterated forcing exercise. 

Let $M$ be a countable transitive model of $ZFC+GCH$ and assume that
  $\kappa<\lambda$ are cardinals with $\aleph _0 <cof(\kappa)$  and
  $\aleph _1<cof(\lambda)$. now i need to force that $2^{\omega}=\kappa$ and
  $2^{\omega _1}=\lambda$ and that all cardinals are preserved.

So i defined $P_1=Fn(\kappa\times\omega,2,\omega)\wedge P_2=Fn(\lambda\times\omega_1,2,\omega_1)$ and i know that $P_1$ force $2^{\omega}=\kappa$ and $P_2$ force $2^{\omega _1}=\lambda$. and then $P=P_2\times P_1$ and we Let $G$ be a $P$-generic filter over $M$, and we know that $G=G_2\times G_1$ for $G_2$- a $P_2$ generic filter over $M$ and $G_1$ a $P_1$ generic filter over $M[G_2]$.
Now i know that $Fn(\kappa\times\omega,2,\omega)^{M[G_2]}=Fn(\kappa\times\omega,2,\omega)^{M}$  does that mean  that forcing with $G_1$ above $M[G_2]$ preserves cardinalities? since $P_1$ has the $c.c.c$ and $2^{{\omega}}$-closed in $M$ so it still has it in $M[G]$ ? is that true?
because if it is true i know how to finish it from there.
thank you
 A: You can show $P_1 \times P_2$ is preserves all cardinals. First, note that since this is a produce forcing (both posets are in the ground model), $M[G_1][G_2] = M[G_2][G_1]$. To show preservation of cardinals, it is easier to work with $P_2 \times P_1$. 
$(2^{<\aleph_1})^{M} = \aleph_1$ since $M$ satisfy the GCH. Therefore, $P_2$ has the $\aleph_2$-c.c. $P_2$ preserve all cardinals at and above $\aleph_2$. Obviously $P_2$ is $(<\aleph_1)$-closed; therefore, $\aleph_1$ is preserved. So $P_2$ preserves all cardinals. $P_1^M = P_1^{M[G]}$ since both models have the same finite partial functions. $P_1$ has the c.c.c. in $M[G_2]$, so $P_1$ preserves all cardinals. Therefore, $M[G_2][G_1] = M[G_1][G_2]$ preserves all cardinals. 
Now to compute exponentials. This time it is easier to work with $P_1 \times P_2$ rather than $P_2 \times P_1$. It is clear that $(2^{\aleph_0})^{M[G_1]} \geq \kappa$. $(2^{\kappa})^{M[G_1]} \leq ((|\kappa \times \omega|^{<\aleph_0})^{\aleph_0})^M = \kappa$ using GCH and the fact that $\aleph_0 < \text{cof}(\kappa)$. Since $P_2$ is $(<\aleph_1)$-closed, it adds no new subsets of $\omega$, so $(2^{\aleph_0})^{M[G_1][G_2]} = (2^{\aleph_0})^{M[G_1]} = \kappa$. Finally, it is clear that $(2^{\omega_1})^{M[G_1][G_2]} \geq \lambda$. By a similar argument to before, you have show it must be $\lambda$. (I think you may need that $\text{cof}(\lambda) > \aleph_1$.)
