24) Complete the proof of the fundamental theorem of arithmetic by proving the uniqueness of the prime factorization up to ordering for positive integers greater than 1.

Pf: Consider $n >1 \in \mathbb{Z}$. We will induct on $n$ to prove that for all $n$, prime factorization is unique. First, consider base case $n=2$. Because 2 is prime, we know that it has exactly two factors, 1 and itself. Thus, the only way to express 2 in the form of the product of primes to positive integer powers is $2^1$. Thus, the base case is proven. Now we may assume that for all $n\leq k$, for some k, prime factorization is unique. We must now simply show that prime factorization is unique for $k+1$. There are two cases. If $k+1$ is prime, prime factorization is clearly unique. If $k+1$ is composite, by definition we may divide it by some prime number to yield another positive integer, $m$, with $m \leq k+1$ and $m > 1$. Since division is well defined, regardless of any chosen factorization of $k+1$, $k+1$ dividing by the same prime always produces the same $m$. And since $m<k$, by inductive hypothesis, $m$ has a unique prime factorization. Thus, $k+1$ has a unique prime factorization. $QED$