n-skeleton:

Show that every positive rational number can be written as a quotient of products of factorials of (not necessarily distinct) primes. For example, \begin{equation*} \frac{10}{9} = \frac{2!\cdot5!}{3!\cdot3!\cdot3!} \end{equation*}

1. A rational number can be expressed as the quotient of two integers.
2. An integer can be expressed as the product of (not necessarily unique) prime numbers.
3. By definition, $n! = n\cdot(n-1)\cdot(n-2)\cdot ... \cdot(1)$. Thus a prime number can be expressed as a quotient of two factorials: $p = \frac{p!}{(p-1)!}$
4. If (p-1) is not prime, express $(p-1)!$ as $(p-1) \cdot (p-2) \cdot ... \cdot (p-n)!$ where $(p-n)$ is prime.
5. Use 2.) to express the $(p-i)$, where $i \in \{1, 2, ..., n-1\}$, as products of primes, and apply 3.)

WLOG, take the example above. $10 = 2 \cdot 5 = \frac{2! \cdot 5!}{1! \cdot 4!} = \frac{2! \cdot 5!}{4 \cdot 3!} = \frac{2! \cdot 5!}{2! \cdot 2! \cdot 3!} = \frac{5!}{2! \cdot 3!}$, $9 = 3 \cdot 3 = \frac{3! \cdot 3!}{2! \cdot 2!}$

$\frac{10}{9} = \frac{5! \cdot 2! \cdot 2!}{2! \cdot 3! \cdot 3! \cdot 3!} = \frac{2! \cdot 5!}{3! \cdot 3! \cdot 3!}$

A $\textit{base 10 over-expansion}$ of a positive integer N is an expression of the form \begin{equation*} N = d_k 10^k + d_{k-1} 10^{k-1} + ... + d_0 10^0 \end{equation*} with $d_k \ne 0$ and $d_i \in \{0,1,2,...,10\} \forall \textit{i}$. For instance, the integer N = 10 has two base 10 over-expansions: $10 = 10 \cdot 10^0$ and the usual base 10 expansion $10 = 1 \cdot 10^1 + 0 \cdot 10^0$. Which positive integers have a unique base 10 over-expansion?