Cardinal number of eventually constant rational sequences If $D_n=\{\langle d_k\rangle\in \mathbb{Q}^\mathbb{N}: (\exists q\in \mathbb{Q})(\forall k\geq n) \,d_k=q\}$, what is the cardinal number of $D_n$?  
Is it $|D_n|=|\mathbb{Q}^\mathbb{N}|=|\mathbb{Q}|^{|\mathbb{N}|}=\aleph_0^{\aleph _0}=\aleph _0$?
 A: Hint: there is a bijection between $D_n$ and $\mathbb{Q}^{n+1}$.
Edit: if you don't agree with me that zero is a natural number, then you will find it easier to construct a bijection between $D_n$ and $\mathbb{Q}^n$.
A: Consider the equivalence relation on $D_n$, $\langle d_k\rangle\equiv\langle d'_k\rangle$ if and only if $d_k=d'_k$ for all $k<n$. This means that if they differ then it's only by the constant end segment of the sequence.
It is not very hard to show $\equiv$ is indeed an equivalence relation on $D_n$. Note that we can also choose from each equivalence class the unique representatives whose constant end segment is all $0$.
This means that there are $|\Bbb Q^n|=|\Bbb Q|^n=\aleph_0^n=\aleph_0$ equivalence classes. Each class contains exactly $|\Bbb Q|=\aleph_0$ different sequences (because each sequence is defined by the initial segment, and the rational from the end segment, and the initial segment is shared by all the sequences, and there are only $\aleph_0$ rational numbers). 
Therefore $|D_n|=\aleph_0\cdot\aleph_0=\ldots$
