How will you prove that there is only one unique way to expand any positive integer factorial in terms of prime factorial numbers?

Question

Ex: Suppose that there exists two ways to write down $x!$, the two ways are $$(2!)^{a2}\times (3!)^{a3} \times (5!)^{a5} \times (7!)^{a7}$$ and $$(2!)^{b2} \times (3!)^{b3} \times (5!)^{b5} \times (7!)^{b7}.$$ How is there only one unique form of expanding $x!$?

Answer 1

The fundamental theorem of arithmetic guarantees a unique prime factorization of $n$, and hence a unique maximal prime that divides $n$, e.g. $p=\max\{p: p|n,p\ \mbox{prime}\}$.

For any given representation $a=(a_1,a_2,\cdots,a_r)$ of $x!$ as you've written, let $p(a_1,a_2,\cdots,a_r)$ denote the maximum prime factorial in the expression. For example if you write $x!=(2!)^{a_2}(7!)^{a_3}$, the maximum prime factorial is $7!$. Since $7$ is prime, it cannot be a factor of any of the lower prime factorials. It also follows that the largest prime that divides $x!$ is this maximal prime. So by the fundamental theorem of arithmetic, the largest prime factorial is unique and is equal to $r!$ . Now just divide both sides by maximal prime factorial, and repeat.