Question on product of primes

Question

How to prove the following result:

$$\prod_{i=1}^{n}P_i=\frac{2^{(P_n+3)/2}}{\sqrt{\pi}} \gamma (1+P_n/2) \cdot \frac1R$$

where $R$ is the product the odd composite natural numbers less than $P_n$ and $n>2$.

Answer 1

Note that $R\prod_1^nP_i$ is just twice the product of all the odd numbers up to $P_n$, and that product of odd numbers is $(P_n+1)!$ divided by the product of the even numbers up to $P_n+1$, and that product in turn is $2^{(P_n+1)/2}$ times the factorial of $(P_n+1)/2$.

Answer 2

Depending on what $\gamma$ is, here is an asymptotic approximation: $$ \begin{align} R\prod_{i=1}^nP_i &=P_n!!\\ &=\frac{(P_n+1)!}{2^{(P_n+1)/2}\left(\frac{P_n+1}{2}\right)!}\\ &\sim\frac1{\sqrt2}\left(\frac{P_n+1}{e}\right)^{\frac{P_i+1}{2}} \end{align} $$