Question 58
For an arbitrary positive integer $s$, we denote by $v_2(s)$ the exponent of the biggest power of $2$ which divides $s$, i.e. $2^{v_2(s)}$ is a divisor of $s$, but $2^{v_2(s)+1}$ is not. Show that for every positive integer $m$ the following equality holds:
$$
v_2\left(\prod_{n=1}^{2^m} \binom{2n}{n}\right) = m2^{m-1} + 1.
$$
$$
v_2\left(\prod_{n=1}^{2^m} \binom{2n}{n}\right) = m2^{m-1} + 1.
$$