Cartesian Product of Countable Sets is Countable/Corollary/Proof 2

Corollary to Cartesian Product of Countable Sets is Countable
Let $k > 1$.

Then the cartesian product of $k$ countable sets is countable.

Proof
Proof by induction:

Base case
When $k = 2$, the case is the same as Cartesian Product of Countable Sets is Countable.

So shown for base case.

Induction Hypothesis
This is our induction hypothesis:
 * $\exists g: S_1 \times S_2 \times \cdots \times S_k \to \N$

Now we need to show true for $n=k+1$:
 * $\exists g: S_1 \times S_2 \times \cdots \times S_k \times S_{k+1} \to \N$

Induction Step
This is our induction step:

Since $S_1 \times S_2 \times \cdots \times S_k$ and $S_{k+1}$ are countable,
 * $\exists f: \left( S_1 \times S_2 \times \cdots \times S_k \right) \times S_{k+1} \to \mathbb N \times \mathbb N$.

By Cartesian Product of Countable Sets is Countable,
 * $\exists r: \mathbb N \times \mathbb N \to \mathbb N$

Therefore, by Composite of Injections is Injection,
 * $\exists g: S_1 \times S_2 \times \cdots \times S_k \times S_{k+1} \to \N$

The result follows by induction.