This inequality is part of a larger probability question and must be shown without explicitly calculating it. The only hint I have is to use Stirling's Approximation so $1 \leq \frac{n!}{\sqrt{2\pi n}(n/e)^n } \leq e^{1/(12n)}$
Edit: As requested the broader question. Proof that for $n=10000$, $p = 0.485$ $$(1-\binom{2n}{n}p^n(1-p)^n\frac{1-p}{1-2p})^{82} \geq 0.99$$ I used Bernoulli's inequality and found and upper bound for $\frac{1-p}{1-2p}$. Note that the original question said 14400 when it should have been 1435
I will only give a very short answer, because it is probably not relevant to anyone else and the solution is actually quite simple. Notice that $\binom{2n}{n} = \frac{2n!}{(n!)^2}$ whose lower bound using Stirling's Approximation is $$\frac{\sqrt{4\pi n}(\frac{2n}{e})^n}{\sqrt{2\pi n}(\frac{n}{e})^n}*e^{1/(24*n)}$$ $$= \frac{1}{\sqrt{\pi n}} 2^{2n}e^{1/(24*n)}$$ From there you can write the whole thing as $$1-1435 \frac{1}{\sqrt{\pi n}} (4p(1-p))^n e^{1/(24*n)}$$ After that you got sane numbers and can do basic estimations.