When [tex]n=4[/tex], you have
[tex]\dbinom84=70>64=2^6[/tex]
but when [tex]n=5[/tex], you have
[tex]\dbinom{10}5=252<256=2^8[/tex]
so [tex]n=5[/tex] is the base case.
Assume the relation holds for [tex]n=k[/tex], i.e.
[tex]\dbinom{2k}k<2^{2k-2}[/tex]
To show that it holds for [tex]n=k+1[/tex], notice that
[tex]\dbinom{2(k+1)}{k+1}=\dfrac{(2k+2)!}{(k+1)!(k+1)!}=\dfrac{(2k+2)(2k+1)(2k)!}{(k+1)^2k!k!}[/tex]
Since [tex]\dbinom{2k}k=\dfrac{(2k)!}{k!k!}[/tex], it must be the case that
[tex]\dbinom{2(k+1)}{k+1}<\dfrac{(2k+2)(2k+1)}{(k+1)^2}2^{2k-2}[/tex]
[tex]\dbinom{2(k+1)}{k+1}<\dfrac{2k+1}{k+1}2^{2k-1}[/tex]
[tex]\dbinom{2(k+1)}{k+1}<\dfrac{k+\frac12}{k+1}2^{2k}<2^{2k}[/tex]
where the last line follows from the fact that the numerator is necessarily smaller than the denominator in [tex]\dfrac{k+\frac12}{k+1}[/tex].
