<s>[INST] How to prove binomial coefficient $ {2^n \choose k} $ is even number? Prove:
${2^n \choose k}$ (for integers $k$ & $n$ : $0<k<2^n$) is even number.
I have tried induction but was unable to get any useful results. 
 [/INST] Provided that you already know $\binom{2^n}{k}$ is an integer, it suffices to show the numerator has more factors of $2$ than the denominator. We have:
$$\binom{2^n}{k}=\frac{(2^n)(2^n-1)\cdots(2^n-k+1)}{k!}.$$
There are at least $n$ factors of $2$ in the numerator because $2^n$ is a factor. So now we need to count the number of factors of $2$ in the denominator, $k!$. 
We have $k!=k(k-1)(k-2)\cdots 3\cdot 2 \cdot 1$. At most $\frac{k}{2}$ of these numbers are divisible by $2$. At most $\frac{k}{4}$ of them are divisible by $4$. At most $\frac{k}{8}$ of them are divisible by $8$. And so on. Each number that is divisible by $2$ contributes a factor of $2$, each number that is divisible by $4$ contributes an additional factor of $2$, each number that is divisible by $8$ contributes a third factor of $2$, and so on. So the number of factors of $2$ in $k!$ is no more than 
$$\frac{k}{2}+\frac{k}{4}+\frac{k}{8}+\cdots = k.$$
Since $k<n$, the denominator has strictly fewer factors of $2$ than the numerator. 
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