Show that an integral domain $A$ is a unique factorization domain if and only if every ascending chain of principal ideals terminates, and every irreducible element of $A$ is prime.
Hint: To show the "if" part, observe that if $a\in A$ is not irreducible, then $(a)\subset (b)$ for some $b\in A$, and continue in this manner to get an ascending chain of principle ideals. Do you see how this gives you a factorization of $a$ into irreducible elements (assuming the chain terminates)? Showing uniqueness is then fairly easy. Suppose $a_1\cdots a_n=b_1\cdots b_m$ where $a_1,\ldots,a_n,b_1,\ldots,b_m$ are prime. Then each $a_i$ divides one of the $b_j$, but these are irreducible so the two must be equal. For the "only if" part, note that every irreducible is prime and factoring an element $a$ into primes gives you all the principle ideals containing $(a)$, of which there are only finitely many.