Prove that if $I$ is uncountable, then $\mathbb{R}^I$ with the product topology is not countable.
Based on what I have read, a set is uncountable if there is a bijection from that set to $\mathbb{N}$. Also, the product topology, according to my textbook is as follows:
Let $(X_1, \mathscr{T}_1) , \ldots , (X_n, \mathscr{T}_n)$ be a finite collection of nonempty topological spaces, and let $X$ denote the Cartesian product $X = \prod_{i=1}^{n} X_i = X_1 \times \ldots \times X_n$.
Let $\mathscr{B}$ be the family of all subsets of $X$ of the form $O = \prod_{i=1}^{n} O_i = O_1 \times \ldots \times O_n$ where each $O_i$ is an open set in the topology $\mathscr{T_i}$ for $X_i$. Then $\mathscr{B}$ is a basis for a topology for $X$ and is called the product topology.
The way I am thinking about this problem is to somehow express $\mathbb{R}$ as a Cartesian product, perhaps as $\mathbb{R}^{1-n} \times \mathbb{R}^n$, and then use the fact that since $\mathbb{R}$ is uncountable by virtue of Cantor's diagonalization argument, and given that $I$ is uncountable, then the product of an uncountable set to the power of another uncountable set must be uncountable.
This last statement, however, sounds more like a play of words more than an actual proof. Could somebody please help me formalize this idea; that is, if it is even the right way to go about solving this problem?
Furthermore, I have seen some posts related to this problem in that they have similar hypotheses, but they aim to show that $\mathbb{R}^I$ is not normal instead of uncountable. There may very well be some relationship between these two properties, but I am not in any position to determine that since I have yet to see this idea of normality covered at all.
As always, any assistance that you provide will be greatly appreciated. Thank you for your time and patience.