Question: Let ${A\_n}$, n = 1, 2, 3, ... be a sequence of countable sets, and put $B = A\_1 \times A\_2 \times \cdots \times A\_n \times \cdots$. Show that B is uncountable. Prove that the same statement holds if each $A\_n$ = {0, 1}. 

I'm not exactly sure how I should begin this. Any help would be appreciated.

Answer: 
Assuming that for you *countable* means *in bijection with the set of natural numbers*, as each $A\_{n}$ is countable, $\vert A\_{n}\vert=\vert\mathbb{N}\vert$. Therefore one has

$$\vert B\vert = \vert\prod\_{n\in\mathbb{N}}A\_{n}\vert = \vert\prod\_{n\in\mathbb{N}}\mathbb{N}\vert =\vert\mathbb{N}^\mathbb{N}\vert\geq\vert 2^{\mathbb{N}}\vert =\vert\mathbb{R}\vert > \vert\mathbb{N}\vert ,$$

where the last equality is due (for example) to the binary representation of real numbers. This chain of inequalities proves both your statements.
$\{0,1\}\times\{0,1\}\times\cdots=\{0,1\}^\mathbb N=2^\mathbb N$ is not countable because no function $f:\mathbb N\rightarrow2^\mathbb N$ is surjective because no $n\in\mathbb N$ maps to $M=\{k\in\mathbb N|k\notin f(k)\}$.

$f(n)=M \Rightarrow (n\in M\Leftrightarrow n\notin M)$.