I am confused with the expression $\dfrac{(N!)^2}{(2N)!}$. Suppose we write $(2N)!=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_\ell^{\alpha_\ell}$ and by Hardy-Ramanujan theorem that states that almost all number $n$ has about $\log\log n$ distinct prime factors.
I want to choose large $N$ for which $\ell<2\log\log (2N)!$. I determine the expression, by letting $\beta$ be such that $p^\beta\le 2N<p^{\beta+1}$ corresponding to each $p_i$, $$\sum_{\alpha\ge 1}\left(2\left\lfloor\dfrac{N}{p^\alpha}\right\rfloor-\left\lfloor\dfrac{2N}{p^\alpha}\right\rfloor \right)=\sum_{1\le\alpha\le\beta}\left(2\left\lfloor\dfrac{N}{p^\alpha}\right\rfloor-\left\lfloor\dfrac{2N}{p^\alpha}\right\rfloor \right)\ge\sum_{\alpha\le \beta}\left(-2\right)=-2\beta.$$
Thus, it means $\dfrac{(N!)^2}{(2N)!}\cdot p_1^{2\beta_1}p_2^{2\beta_2}\dots p_\ell^{2\beta_\ell}\in\mathbb Z_{>0}$. However,
\begin{align*} \dfrac{(N!)^2}{(2N)!}\cdot p_1^{2\beta_1}p_2^{2\beta_2}\dots p_\ell^{2\beta_\ell} & \le\dfrac{(N!)^2}{(2N)!}\cdot (2N)^{2\ell}\\ & <\dfrac{(N!)^2}{(2N)!}\cdot (2N)^{4\log\log (2N)!}\\ & <\dfrac{(N!)^2}{(2N)!}\cdot (2N)^{4\log(2N\log(2N))} \end{align*}
and the last one tends to $0$ since $\displaystyle \dfrac{(N!)^2}{(2N)!}\ll \dfrac{\sqrt{N}}{4^N}$ by Stirling approximation, and $\sqrt{N}\cdot(2N)^{4\log(2N\log(2N))}=\text o(4^N)$ because $\log\sqrt{N}+4\log(2N)(\log(2N)+\log\log(2N))=\text o(N)$.
How is this possible? What is my mistake for the argument? Thank you very much in advance.