Putnam 2009 B1 (rational number as factorial) 
Show that every positive rational number can be written as a quotient of products of factorials of (not necessarily distinct) primes. For example, $ \frac{10}9=\frac{2!\cdot 5!}{3!\cdot 3!\cdot 3!}.$

I used the idea:
$$\frac{a}{b} = \frac{\prod x_{k}!}{\prod y_{j}!} \implies a = \frac{(b)\cdot\prod x_{k}!}{\prod y_{j}!}$$
Then I said suppose it holds for:
$$\varphi = \{1, 2, 3.... a-1\}$$
Then since for $a > 1$:
$$\frac{a}{2} < a-1$$ It follows from strong induction,
$$\frac{a}{2!} = \frac{(b)\cdot\prod x_{k}!}{\prod y_{j}!}$$
$$\frac{a}{b} = 2! \cdot \frac{ \prod x_{k}!}{\prod y_{j}!}$$

Proposition. Every prime number can be written as quotient of prime factorials (pfq for short).
Proof. We have
 $2=2!$ (or $2=\frac{2!97!}{97!}$ if you insist on a denominator).
Let $p>2$ be a prime and assume all primes $<p$ can be written as pfq. Since $p=\frac{p!}{(p-1)!}$ and $(p-1)!$ is a product of primes $<p$, we conclude that the denominator and ultimately $p$ can also be written as pfq.
$_\square$
Corollary. The original claim.
Proof. Just write down the prime factorization of numerator and denominator and use the proposition for each prime. $_\square$
