Let $J$ be some (non-empty) set of indices, and let $\left\{ A_\alpha \right\}_{\alpha \in J}$ and $\left\{ B_\alpha \right\}_{\alpha \in J}$ be any two indexed families of sets such that, for each $\alpha \in J$, the sets $A_\alpha$ and $B_\alpha$ have the same cardinality (i.e. are numerically equivalent). Then do the Cartesian products $\prod_{\alpha \in J} A_\alpha$ and $\prod_{\alpha \in J} B_\alpha$ also have the same cardinality?
My Attempt:
The answer is yes. The proof goes as follows:
For each $\alpha \in J$, as $A_\alpha \sim B_\alpha$, so we can find a bijective function $f_\alpha \colon A_\alpha \longrightarrow B_\alpha$. Now let $\mathbf{f} \colon \prod_{\alpha \in J} A_\alpha \longrightarrow \prod_{\alpha \in J} B_\alpha$ be the function defined by the formula $$ \mathbf{f} \left( \left( a_\alpha \right)_{\alpha \in J} \right) := \left( f_\alpha \left( a_\alpha \right) \right)_{\alpha \in J}. $$
We show that this function $\mathbf{f}$ is also bijective.
Let $\left( a_\alpha \right)_{\alpha \in J}$ and $\left( a_\alpha^\prime \right)_{\alpha \in J}$ be any two $J$-tuples (i.e. elements) of $\prod_{\alpha \in J} A_\alpha$ such that $$ \mathbf{f} \left( \left( a_\alpha \right)_{\alpha \in J} \right) = \mathbf{f} \left( \left( a_\alpha^\prime \right)_{\alpha \in J} \right), $$ that is, such that $$ \left( f_\alpha \left( a_\alpha \right) \right)_{\alpha \in J} = \left( f_\alpha \left( a_\alpha^\prime \right) \right)_{\alpha \in J}. $$ Then, for each $\alpha \in J$, we have $$ f_\alpha \left( a_\alpha \right) = f_\alpha \left( a_\alpha^\prime \right), $$ and since $a_\alpha, a_\alpha^\prime \in A_\alpha$ and since the function $f_\alpha \colon A_\alpha \longrightarrow B_\alpha$ is injective, therefore we can conclude that $a_\alpha = a_\alpha^\prime$ for each $\alpha \in J$; hence we have $$ \left( a_\alpha \right)_{\alpha \in J} = \left( a_\alpha^\prime \right)_{\alpha \in J}, $$ thus showing that $\mathbf{f}$ is injective.
Let $\mathbf{b} \colon= \left( b_\alpha \right)_{\alpha \in J}$ be any element of $\prod_{\alpha \in J} B_\alpha$. We show that there exists an element $\mathbf{a} \in \prod_{\alpha \in J} A_\alpha$ such that $$ \mathbf{b} = \mathbf{f} ( \mathbf{a} ). $$
For each $\alpha \in J$, as $b_\alpha \in B_\alpha$ and as the function $f_\alpha \colon A_\alpha \longrightarrow B_\alpha$ is surjective, so there exists an element $a_\alpha \in A_\alpha$ such that $$ b_\alpha = f_\alpha \left( a_\alpha \right). $$ Let us put $$ \mathbf{a} := \left( a_\alpha \right)_{\alpha \in J}. $$ This $\mathbf{a} \in \prod_{\alpha \in J} A_\alpha$ of course, and furthermore we also have $$ \mathbf{f} ( \mathbf{a} ) = \mathbf{b}, $$ thus showing that $\mathbf{f}$ is surjective.
Thus we have shown that $$ \prod_{\alpha \in J} A_\alpha \sim \prod_{\alpha \in J} B_\alpha. $$
Is this proof correct and clear enough? If so, then how about the following generalisation of the above result?
Let $I$ and $J$ be any two (non-empty) sets of indices such that $I \sim J$, and let $\left\{ A_\alpha \right\}_{\alpha \in I}$ and $\left\{ B_\beta \right\}_{\beta \in J}$ be any two indexed families of sets such that, for each $\alpha \in I$, we have $A_\alpha \sim B_{\varphi(\alpha)}$, where $\varphi \colon I \longrightarrow J$ is a bijective function. Then $$ \prod_{\alpha \in I} A_\alpha \sim \prod_{\beta \in J} B_\beta. $$
Does the result as I have just stated also hold?