The Fundamental Theorem of Arithmetic

We can now prove Theorem 1.3. The idea is simple. Suppose we have two factorization. Use Theorem 2.4 to cancel primes from each, one prime at a time. At the end of the game, we discover that the factorizations have to consist of exactly the same primes. The technical details, with all the $p$'s and $q$'s are given below:

Proof. We have

$$n = p_1\cdot p_2 \cdots p_d,$$

with each $p_i$ prime. Suppose that

$$n = q_1\cdot q_2 \cdots q_m$$

is another expression of $n$ as a product of primes. Since

$$p_1 \mid n = q_1\cdot (q_2 \cdots q_m),$$

Euclid's theorem implies that $p_1 = q_1$ or $p_1 \mid q_2\cdots q_m$. By induction, we see that $p_1 = q_i$ for some $i$.

Now cancel $p_1$ and $q_i$, and repeat the above argument. Eventually, we find that, up to order, the two factorizations are the same. $\qedsymbol$