Is the cardinality of the infinite sequences consisting of $\left\{0,1\right\}$ or $\left\{0,1,2\right\}$ or $\left\{0,1,2,3,\cdots,9 \right\}$ elements equal to the cardinality of the infinite sequences consisting of elements $\left\{0,1\right\}$?
I mean, is this correct?
$$card \left(2^{\mathbb{N}}\right)=card \left(3^{\mathbb{N}}\right)=card \left(4^{\mathbb{N}}\right)=\cdots=card \left(9^{\mathbb{N}}\right)=2^{\aleph_0}$$
If correct, where can I find proof of that?
Yes, $\alpha^{\aleph_0}=2^{\aleph_0}$ for all $2\le\alpha\le 2^{\aleph_0}$. This is due to the identity $\aleph_0^2=\aleph_0$ and inequalities $$2^{\aleph_0}\le \alpha^{\aleph_0}\le \left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0^2}=2^{\aleph_0}$$