Solve $C_N = (2 + 1/\log_2N)C_{N/2}$ for $N \ge2$ and $C_1 = 1$

Question

Solve $C_N = (2 + 1/\log_2N)C_{N/2}$ for $N \ge2$ and $C_1 = 1$.

My solution: $N = 2^n$

$C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$

$C_N = 3*(5/2)*...*(2+1/lgN)$

How can I simplify further?

Answer 1

$C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$
$(2 + 1/n) = (2n + 1)/n$
or
$C_{2^n} = (3 * 5 * 7 * ... * (2n + 1)) / (1 * 2 * 3 ... * n)$
or
$C_{2^n} = (2n + 1)! / ((2^n)* n! * n!)$