$\operatorname{card}(\bigcup\limits_{n \in \mathbb{N}} \underbrace{A\times…\times A}_{n})=k$ if $\operatorname{card}(A)=k$ infinite.

I was reading a proof of a theorem that goes like this:
Let $$A$$ be an infinite set of cardinality $$k$$ and $$A^{<\omega}$$ the set of finite sequences of elements of A. Then $$\operatorname{card}(A^{<\omega})=k$$.

In the proof, we note that $$A^{<\omega}=\bigcup\limits_{n \in \mathbb{N}} \underbrace{A\times...\times A}_{n}$$. So at some point, we have to show $$\operatorname{card}(A \times A)=k$$, and for this the author uses the fact that $$\operatorname{card}(A \times A)=\operatorname{card}(\operatorname{card}(A) \times \operatorname{card}(A))$$.
I am not super sure why this is the case. My thoughts: We have by definition a bijection $$A \leftrightarrow \operatorname{card}(A)$$. So we have a bijection $$A \times A \leftrightarrow \operatorname{card}(A) \times \operatorname{card}(A)$$. So in particular since these sets are bijective they have the same cardinal $$\operatorname{card}(A \times A)=\operatorname{card}(\operatorname{card}(A) \times \operatorname{card}(A))$$. Is that it?
So from this we can prove $$\operatorname{card}(\underbrace{A\times...\times A}_{n})=k$$ $$\forall n$$.

Then the author says $$k\le \operatorname{card}(\bigcup\limits_{n \in \mathbb{N}} \underbrace{A\times...\times A}_{n}) \le \operatorname{card}(A)\cdot\omega$$ where the $$\cdot$$ is the cardinal product. Why is that?

So far, so good. Next you need $$k^2=k$$ to prove by induction that each of the products has cardinality $$k$$. Then their union has at most size $$k\aleph_0\le k^2=k$$. But the proof there are at least $$k$$ sequences is trivial.
• In general, what bound do we have : $\operatorname{card}(\bigcup\limits_{i \in I} B_i) \le ?$ – roi_saumon Jan 23 at 23:56
• @rol_saumon We have lower bound $\max_i\operatorname{card}(B_i)$, upper bound $\sum_i\operatorname{card}(B_i,)$. – J.G. Jan 24 at 6:02