Proving $\mathbb{N}^k$ is countable Prove that $\mathbb{N}^k$ is countable for every $k \in \mathbb{N}$.
I am told that we can go about this inductively. 
Let $P(n)$ be the statement: “$\mathbb{N}^n$ is countable” $\forall n \in \mathbb{N}$.
Base Case: $\mathbb{N}^1 = \mathbb{N}$ is countable by definition, so $\checkmark$
Inductive Step: $\mathbb{N}^{k+1}$ $“=”$ $\mathbb{N}^k \times \mathbb{N}$
We know that $(A,B)$ countable $\implies$ $A \times B$ is countable. I am stuck on the part where I have to prove the rest, but I know that, for example, $(1,2,7) \in \mathbb{N}^3 \notin \mathbb{N}^2 \times \mathbb{N}$ but instead $((1,2),7) \in \mathbb{N}^2 \times \mathbb{N}$. So how would I go about proving the statement.

The function $f:\mathbb{N^K}\to \mathbb{N^K\times \{m \}}$ defined by $$f(a_1,a_2,\cdots, a_k)=(a_1,a_2,\cdots, a_k,m)$$ is clearly a bijection for fixed $m\in \mathbb{N}$ and we can write $\mathbb{N^{K+1}}$ as $$\mathbb{N^{K+1}}=\bigcup_{m=1}^{\infty}\{\mathbb{N^K\times \{m \}}\}$$ and this being a countable union of countable sets is countable.
