Find all natural numbers $n > 1$ and $m > 1$ such that $1!3!5!\cdots(2n - 1)! = m!$ 
Find all natural numbers $n > 1$ and $m > 1$ such that $1!3!5!\cdots(2n - 1)! = m!$

I have been thinking about coming up with some inequalities which would narrow the possible range of pairs $(n, m)$, however the best I have been able to find so far are $m \ge 2n - 1$ and $m \lt n^2$, which is clearly not enough.
Update.
The question has tag combinatorics as it is from a book about combinatorics, so there must be at least partly combinatorial solution.
 A: Note that every prime divisor of $1!3!5!\cdots(2n-1)!$ occurs at least twice, except for perhaps $2n-1$ (if it's prime). By Bertrands Postulate, we already know that $m!$ has one prime divisor that occurs only once. (See Can n! be a perfect square when n is an integer greater than 1?) A slightly stronger version of the theorem will give that there are at least two primes in $m!$ with multiplicity $1$. Indeed, Wikipedia says that for $x\geq25$ there are at least two primes between $x$ and $(1+0.2)^2x=1.44x$. Thus there are no solutions for $m\geq50$ (and you might even reduce this upper bound by checking some cases by hand - that is, looking for $m$ such that there are at least two prime divisors of $m!$ with multiplicity one).
The only solutions are $(2,3),(3,6),(4,10)$.
A: Let
$$N:=m! =1!\>3!\>5!\cdots(2n-1)!$$
for certain numbers $m$, $n\in{\mathbb N}_{\geq1}$, and denote by $p$ the exponent of $2$ in the prime decomposition of $N$. Then one has on the one hand
$$p=\left\lfloor{m\over2}\right\rfloor+\left\lfloor{m\over4}\right\rfloor+\ldots <m$$
and on the other hand
$$p\geq\sum_{k=1}^n\left\lfloor{2k-1\over2}\right\rfloor={(n-1)n\over 2}\ .$$
This implies
$$m>{(n-1)n\over2}\ .$$
On the other hand, by Bertrand's postulate, we must have $m<2(2n-1)$, because otherwise $m!$ would contain a prime factor not present in $(2n-1)!$. Now
$${(n-1)n\over2}<2(2n-1)$$
enforces $n\leq8$. But we can do better, since for small values of $2n-1$ there are many more primes available than Bertrand's postulate guarantees. We therefore set up the following table:
$$\matrix{n&&1&2&3&4&5&6&7&8\cr
2n-1&&1&3&5&7&9&11&13&15\cr
m>&&0&2&3&6&10&15&21&28\cr}$$
This table show that already for $n\geq5$ any "admissible" $m!$ would contain a prime factor $>2n-1$. So it remains to check the cases $n\in[4]$, which lead to
$$N\in\{1, \>6, \>720, \>3628800\}=\{1!,\>3!,\>6!,\>10!\}\ .$$
It follows that there are exactly $4$ pairs $(m,n)$ of the required kind.
