Since $a_i! <(a_1+a_2+...+a_n)! \forall i=\overline{1,n}$, every prime divisor of the denominator is also divides the numerator. Let $p$ be a prime divisor of $(a_1+a_2+...+a_n)!!$ By Legendre's formula we have $v_p((a_1+a_2+...+a_n)!!)=\displaystyle \sum _{i=1}^{\infty} [\dfrac{(a_1+a_2+...+a_n)!}{p^i}]$ $v_p(a_1!!a_2!!\cdots a_n!!)=\displaystyle \sum _{i=1}^{\infty} \left( \displaystyle \sum _{j=1}^{n} [\dfrac{a_1!}{p^i}] \right)$ Since $(a_1+a_2+...+a_n)!>a_1!+a_2!+...+a_n!$, we have $[\dfrac{(a_1+a_2+...+a_n)!}{p^i}] >\displaystyle \sum _{j=1}^{n} [\dfrac{a_1!}{p^i}] \forall i \geq 1$ $\Rightarrow v_p((a_1+a_2+...+a_n)!!)>v_p(a_1!!a_2!!\cdots a_n!!)$ for all prime divisor $p$ of the numerator. Therefore, $\frac{(a_1+a_2+\cdots+a_n)!!}{a_1!!a_2!!\cdots a_n!!}$ is an integer.