Fundamental Theorem of Arithmetic

Theorem
Every natural number $$n>1$$ can be factored into a unique product of primes.

Proof
If $$n$$ is prime, the result is immediate, so we assume $$n$$ is composite. This being the case, the set $$\{d:$$ $$d|n$$ and $$1n_1>1$$. If $$n_1$$ is prime, we are done. Otherwise, the set $$\{d:$$ $$d|n_1$$ and $$1n_1>n_2> \dots >1$$ is a decreasing sequence of positive integers, there must be a finite number of $$n_i$$'s. That is, we will arrive at some prime number $$n_{k-1}$$, which we will call $$p_k$$. This results in the prime factorization $$n={p_1}{p_2} \dots {p_k}$$.

Now, suppose $$n$$ has two prime factorizations, namely $$n=p_1 p_2 \dots p_r=q_1 q_2 \dots q_s$$, where $$r \le s$$ and each $$p_i$$ and $$q_j$$ is prime with $$p_1 \le p_2 \le \dots \le p_r$$ and $$q_1 \le q_2 \le \dots \le q_s$$. Since $$p_1|q_1 q_2 \dots q_s$$, it must be the case that $$p_1=q_j,$$ for some $$1 \le j \le s$$. This means $$p_1 \ge q_1$$. Similarly, since $$q_1|p_1 p_2 \dots p_r$$, we have $$q_1 \ge p_1$$. Thus, $$p_1=q_1$$, so we may cancel these common factors, which gives us $${p_2}{p_3} \dots {p_r}={q_2}{q_3} \dots {q_s}$$.

We may repeat this process to show that $$p_2=q_2$$, $$p_3=q_3$$, $$\dots$$, $$p_r=q_r$$. If $$r<s$$, we arrive at $$1={q_{r+1}}{q_{r+2}} \dots q_s$$ after canceling all common factors. Of course, this is impossible, which means $$r=s$$. Thus, $$p_1=q_1$$, $$p_2=q_2$$, $$\dots$$, $$p_r=q_s$$, which means the two factorizations are identical. Therefore, the prime factorization of $$n$$ is unique.