As this question clearly shows that the countable many product of countable sets is uncountable. However, I do not understand why the below proof is wrong:
(False) Proof:
Let $A$ be a countable set. We use induction to show that the countably many product of $A$ with itself is countable. We use induction. When $n=1$, the theorem is true by our hypothesis.Let assume it is true for $A^n = A\times ... A : n$ times.
Since $A^{n+1} = A^n \times A$, which is the finite product of countable sets, it is also countable by this question. Hence, by induction $A^n$ is countable for all $n \in \mathbb{N}$. QED
Question:
Why is the above proof wrong ? Where is the flaw ?
Edit:
I'm trying to show that $A^{|\mathbb{N}|}$ is countable.
The proof you give proves that $A^n$ is countable for all $n \in \mathbb{N}$ however what you wish to prove is $A^\omega$ is countable. That is the propitiation is true for the infinite case. However $\omega$ isn't in $\mathbb{N}$ so the induction you use doesn't work.