Integer is Expressible as Product of Primes/Proof 3

Proof
The proof proceeds by induction.

For all $n \in \N_{> 1}$, let $P \left({n}\right)$ be the proposition:
 * $n$ can be expressed as a product of prime numbers.

First note that if $n$ is prime, the result is immediate.

Basis for the Induction
$P \left({2}\right)$ is the case:
 * $n$ can be expressed as a product of prime numbers.

As $2$ itself is a prime number, and the result is immediate.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({j}\right)$ is true, for all $j$ such that $2 \le j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * For all $j \in \N$ such that $2 \le j \le k$, $j$ can be expressed as a product of prime numbers.

from which it is to be shown that:
 * $k + 1$ can be expressed as a product of prime numbers.

Induction Step
This is the induction step:

If $k + 1$ is prime, then the result is immediate.

Otherwise, $k + 1$ is composite and can be expressed as:
 * $k + 1 = r s$

where $2 \le r < k + 1$ and $2 \le s < k + 1$

That is, $2 \le r \le k$ and $2 \le s \le k$.

Thus by the induction hypothesis, both $r$ and $s$ can be expressed as a product of primes.

So $k + 1 = r s$ can also be expressed as a product of primes.

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Second Principle of Mathematical Induction.

Therefore, for all $n \in \N_{> 1}$:
 * $n$ can be expressed as a product of prime numbers.