Why does the square root of n! grow exponentially faster than exponential functions?

I am going through Normal Subgroup Reconstruction and Quantum Computation Using Group Representations by Hallgren et al.

In the proof of the theorem $6$ of the paper on page 632, the authors go on proving the difference between the probabilities of sampling all irreps, $|p - q|_1$ of a subgroup inside the symmetric group $S_n$.

$$
\begin{align}
|p - q|_1 &= \Sigma_{\rho} \mid p_{\rho} - q_{\rho} \mid \\
&\le \Sigma_{\rho} \frac{d_{\rho}}{n!} 2^{O(n)} \sqrt{n}^{n / 2} \\
&\le \Sigma_{\rho} \frac{\sqrt{n!}}{n!} 2^{O(n)} \sqrt{n}^{n / 2} \\
&\le \frac{2^{O(n)} \sqrt{n}^{n/2}}{\sqrt{n!}} \\
&= 2^{O(n)} \frac{\sqrt{\sqrt{n}^n}} {\sqrt{n!}} \\
&\le 2^{O(n)} \frac{1}{\sqrt{\left( n / 2 \right)!}} \lll 2^{-\Omega(n)}
\end{align}
$$

How is $2^{O(n)} \frac{1}{\sqrt{\left( n / 2 \right)!}} \lll 2^{-\Omega(n)}$?

$\log(n!)=\Theta(n\log n)$, (see Stirling's approximation), hence

$\frac{2^{O(n)}}{\sqrt{\left(\frac{n}{2}\right)!}}=2^{O(n)-\log\sqrt{\left(\frac{n}{2}\right)!}}=2^{O(n)-\Theta(n\log n)}=2^{-\left(\Theta(n\log n)-O(n)\right)}=2^{-\Omega(n)}$.

StackExchange Computer Science Q#56318, answer score: 9