Assert the range of a binomial coefficient divided by power of a number

Write the expression as $$frac {200!}{(2^{100}cdot 100!)(2^{100}cdot 100!)}$$ The first term in the denominator cancels all the even factors in the numerator, leaving us with $$left(frac{199}{200}right)left(frac{197}{198}right)left(frac{195}{196}right)ldotsleft(frac{1}{2}right)$$ The last factor is $frac 12$ and all the rest are less than $1$, so the product is less than $frac 12$

$$binom{n}{0}+binom{n}{1}+binom{n}{2}+...+binom{n}{n}=2^n$$ so you can rewrite $$begin{equation*} frac{200 choose 100}{4^{100}}=frac{200 choose 100}{2^{200}}=frac{200 choose 100}{binom{200}{0}+binom{200}{1}+binom{200}{2}+...+binom{200}{100}+...+binom{200}{199}+binom{200}{200}}<<1 end{equation*}$$ you can make a sence by looking at pascal's triangle $$1\1space 1\1 space 2 space 1 \ 1 space 3 space 3 space 1\1space 4space 6 space 4 space 1\1space 5 space 10 space 10 space 5space 1\1space 6 space 15 space 20 space 15space 6 space 1$$ for example at $n=4to frac{6}{16}<frac8{16}$ and $n=6to frac{20}{64}<frac 13$ and so on

Hint:

Compare $binom{200}{100}$ with $sum_{k=0}^{200}binom{200}k$.

This in the understanding that: $$sum_{k=0}^{200}binom{200}k=2^{200}=4^{100}$$