prove that for all $n\ge 1$, $a_n/b_n$ is of the form $a/2^b$ where $a$ is an odd integer and $b < 2n$ 
For any $n\ge 1,$ let $a_n = 1\cdot 3\cdots (2n-1)$ and let $b_n = 2\cdot 4\cdots (2n)$. Prove that for all $n\ge 1$, $a_n/b_n$ is of the form $a/2^b$ where $a$ is an odd integer and $b < 2n$.

I initially thought of a proof by induction but it seems hard to come up with the inductive hypothesis. In particular, I initially thought there was some formula for $a$ and/or $b$ in terms of n. Let $c_n$ and $d_n$ be the unique positive integers so that $\dfrac{a_n}{b_n} = \dfrac{c_n}{d_n}$ where $c_n$ is odd and $d_n < 2^2n$ is a power of two. Computation gives $(c_n, d_n) = (1,2), (3,8),(5,16), (35, 128).$ Assume the result holds for $n$. We want to show it holds for $n+1$. In the inductive step, we have $\dfrac{c_{n+1}}{d_{n+1}} = \dfrac{c_n}{d_n} \cdot \dfrac{2n+1}{2n+2}$. It's crucial that the greatest odd divisor of $2n+2$ divide $c_n,$ since all these factors are coprime to $2n+1$ and if the greatest odd divisor of $2n+2$ doesn't divide $c_n$, then there's a prime number occurring to a greater exponent in the product than in $c_n$, and so this prime power will divide $d_{n+1}$.
 A: As TheBestMagician noted, $a_n = (2n-1)!!$ and $b_n = (2n)!! = 2^n n!$.  We may write the central binomial coefficient as
$$\binom{2n}{n} = \frac{(2n)!}{n!^2} = \frac{(2n)!!(2n-1)!!}{n!^2} = \frac{2^n (2n-1)!!}{n!}.$$  Therefore,
$$\frac{a_n}{b_n} = \frac{(2n-1)!!}{(2n)!!} = \frac{(2n-1)!!}{2^n n!} = \frac{1}{2^{2n}} \binom{2n}{n}.$$
The desired result holds provided $\binom{2n}{n}$ is even for $n \ge 1$.  This is true because for $n \ge 1$ we can organize the subsets of $n$ objects out of $2n$ objects into complementary pairs (e.g., for $n = 2$ these pairs are $\{\{1,2\},\{3,4\}\}$, $\{\{1,3\},\{2,4\}\}$, and $\{\{1,4\},\{2,3\}\}$). In fact, the exact value of $b$ in the original problem statement is $2n$ minus the number of ones in the binary representation of $n$.
A: Unless I'm being silly, because $a_n$ is always odd, we just want to show that $\nu_2(b_n)<2n$ for all $n$. This isn't hard because $b_n=2\cdot 4\cdots 2n=2^n\cdot n!$ so $\nu_2(b_n)=n+\nu_2(n!)$. It suffices to show $\nu_2(n!)<n$ for all $n$, which is well-known and easy to prove, by say Legendre's formula.
