<s>[INST] show $k<2^m$ if $2^{2^m}+1=\binom{n}{k}$ let $m$ be give positive integer,and $n,k$ be positive integer ,such $n\ge 2k\ge 2$,and
$$2^{2^m}+1=\binom{n}{k}$$
show that: $k<2^m$
Some of my ideas and attempts:
$$\binom{n}{k}=\dfrac{n!}{k!(n-k)!}=\dfrac{n(n-1)(n-2)\cdots(n-k+1)}{k!}=\left(1+\dfrac{n-k}{k}\right)\left(1+\dfrac{n-k}{k-1}\right)\cdots\left(1+\dfrac{n-k}{1}\right)\ge (1+\dfrac{n-k}{\sqrt[k]{k!}})^k\ge\left(1+\dfrac{k}{\sqrt[k]{k!}}\right)^k\ge 1+\dfrac{k^2}{\sqrt[k]{k!}}$$
at last two step use Holder inequality and  Bernoulli’s inequality
so we have
$$2^{2^m}\ge \dfrac{k^2}{\sqrt[k]{k!}}$$Now, I can’t prove it $k<2^m$
 [/INST] If $k = 1$, clearly, $k < 2^m$.
In the following, assume that $k \ge 2$.
We have, for all $q > k \ge 0$,
$$\binom{q + 1}{k} = 
\frac{(q + 1)!}{k! (q + 1 -k)!} = \frac{q!}{k! (q - k)!} \frac{q + 1}{q + 1 - k}
\ge \binom{q}{k}.$$
Thus, we have
$$\binom{n}{k} \ge \binom{2k}{k}.$$
Using Mathematical Induction, it is easy to prove that, for all $k \ge 2$,
$$\binom{2k}{k} > 2^k + 1.$$
(Note: Use $\binom{2k+2}{k+1} = \binom{2k}{k}\frac{(2k+1)(2k+2)}{(k+1)^2} > 2\binom{2k}{k}$.)
Thus, we have
$$\binom{n}{k} > 2^k + 1.$$
Thus, we have
$$2^{2^m} + 1 > 2^k + 1$$
which results in
$k < 2^m$.
We are done.
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