$(4)^n$> ${2n\choose n}$
I have attempted to prove this by placing it as $(4)^n$=$(1+1)^n$ with another square like $2^{2n},$ since I can't write it properly; then I used the formula for ${2n\choose n}$ which I separated into a sum, ${2n\choose 0}$ + ${2n\choose 1}$ + ... + ${2n\choose n}$, but I still came to the wrong conclusion that $(4)^n$>$(4)^n$, so could you please lend a hand?
Your inequalities are strange... $(1+1)^n=4^n$ is of course wrong. But the idea is good : $$4^n=(1+1)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k},$$ and thus $4^n>\binom{2n}{n}$. By the way, this also prove that $$4^n\geq \sum_{k=m}^{p}\binom{2n}{k},$$ for all $0\leq m\leq p\leq 2n.$