induction proof of fundamental theorem of arithmetic

We present an induction proof by Zermelo for thefundamental theorem of arithmetic (http://planetmath.org/FundamentalTheoremOfArithmetic).

Part 1.  Every positive integer $n$ is a product of prime numbers.

Proof.  If $n=1$, it is the empty product of primes, andif $n=2$, it is a prime number.

Let then $n>2$.  Make the induction hypothesis that all positiveintegers $m$ with    are products of prime numbers. If $n$ is a prime number, the thing is ready.  Else, $n$ is aproduct of smaller numbers; these are, by the induction hypothesis,products of prime numbers.  The proof is complete.

Part 2.  For any positive integer $n$, its representationas product of prime numbers is unique up to the order of theprime factors.

Proof.  The assertion is clear in the case that $n$ is aprime number, especially when  $n=2$.

Let then  $n>2$ and suppose that the assertion is true forall positive integers less than $n$.

If now $n$ is a prime, we are ready.  Therefore let it be acomposite number.  There is a least nontrivial factor $p$ of $n$.This $p$ must be a prime.  Put  $n=pb$  where  b is apositive integer.  By the induction hypothesis, $b$ has aunique prime factor decomposition.  Thus $n$ has a unique primedecomposition containing the prime factor $p$.

Now we will show that $n$ cannot have other prime decompositions. Make the antithesis that $n$ has a different prime decomposition;let $q$ be the least prime factor in it.  Now we have   and  $n=qc$  where  $c\in {\mathbb{Z}}_{+}$  and   with  $p\nmid c$.  Then

$n_0:=n-pc=\{\begin{array}{cc}pb-pc=p(b-c)\hfill & \\ qc-pc=(q-p)c\hfill & \end{array}$

is a positive integer less than $n$.  Since  $p\mid n_0$, theinduction hypothesis implies that the prime $p$ is in the primedecomposition of $(q-p)c$ and thus also at least of $q-p$or $c$.  But we know that  $p\nmid c$, whence  $p\mid q-p$. Thus we would get  $p\mid q-p+p=q$.  Because both $p$and $q$ are primes, it would follow that $p=q$.  Thiscontradicts the fact that  .  Consequently, ourantithesis is wrong.  Accordingly, $n$ has only one primedecomposition, and the induction proof is complete.