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. 

The following solution (copypasted from my old coursework) is purely
elementary number theory:
We set $\mathbb{N}=\left\{  0,1,2,\ldots\right\}  $.

Lemma 1. Let $n$ be an integer. Let $m$ be a positive integer. Then,
  $\dbinom{n}{m}=\dfrac{n}{m}\cdot\dbinom{n-1}{m-1}$.

Proof of Lemma 1. We have
\begin{align*}
\dbinom{n}{m} &  =\dfrac{n\cdot\left(  n-1\right)  \cdot\cdots\cdot\left(
n-m+1\right)  }{m!}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\dbinom{n}{m}\right)  \\
&  =\dfrac{n\cdot\left(  \left(  n-1\right)  \cdot\left(  n-2\right)
\cdot\cdots\cdot\left(  n-m+1\right)  \right)  }{m\cdot\left(  m-1\right)
!}\\
&  \qquad\qquad\left(
\begin{array}
[c]{c}
\text{since}\\
n\cdot\left(  n-1\right)  \cdot\cdots\cdot\left(  n-m+1\right)  =n\cdot\left(
\left(  n-1\right)  \cdot\left(  n-2\right)  \cdot\cdots\cdot\left(
n-m+1\right)  \right)  \\
\text{and }m!=m\cdot\left(  m-1\right)  !
\end{array}
\right)  \\
&  =\dfrac{n}{m}\cdot\dfrac{\left(  n-1\right)  \cdot\left(  n-2\right)
\cdot\cdots\cdot\left(  n-m+1\right)  }{\left(  m-1\right)  !}\\
&  =\dfrac{n}{m}\cdot\underbrace{\dfrac{\left(  n-1\right)  \cdot\left(
n-2\right)  \cdot\cdots\cdot\left(  \left(  n-1\right)  -\left(  m-1\right)
+1\right)  }{\left(  m-1\right)  !}}_{\substack{=\dbinom{n-1}{m-1}
\\\text{(since this is how }\dbinom{n-1}{m-1}\text{ is defined)}}}\\
&  \qquad\qquad\left(  \text{since }n-m=\left(  n-1\right)  -\left(
m-1\right)  \right)  \\
&  =\dfrac{n}{m}\cdot\dbinom{n-1}{m-1}.
\end{align*}
Thus, Lemma 1 is proven. $\blacksquare$

Lemma 2. Let $x$, $y$ and $z$ be three integers such that $x\mid yz$ and
  $\gcd\left(  x,y\right)  =1$. Then, $x\mid z$.

Lemma 2 is a classical result in elementary number theory (see, e.g.,
Proposition 1.2.8 in my 18.781 (Spring 2016): Floor and arithmetic
functions).
$\blacksquare$

Lemma 3. Let $p$ be a prime number. Then, every positive divisor of
  $p^{\alpha}$ is a power of $p$.

Proof of Lemma 3. Let $d$ be a positive divisor of $p^{\alpha}$. We must
show that $d$ is a power of $p$.
Assume the contrary. Thus, the prime factorization of $d$ must contain at
least one prime $q$ distinct from $p$. Consider this $q$. Now, $q\mid d$
(since the prime factorization of $d$ contains $q$). Hence, $q\mid d\mid
p^{\alpha}$ (since $d$ is a divisor of $p^{\alpha}$). Thus, the prime
factorization of $p^{\alpha}$ contains the prime $q$ (since $q$ is a prime).
Since this prime factorization is clearly $\underbrace{pp\cdots p}
_{\alpha\text{ times}}$, we thus conclude that the prime factorization
$\underbrace{pp\cdots p}_{\alpha\text{ times}}$ contains $q$. Hence, $q=p$.
This contradicts the fact that $q$ is distinct from $p$. This contradiction
proves that our assumption was wrong; hence, Lemma 3 is proven. $\blacksquare$

Lemma 4. Let $p$ be a prime number. Let $\alpha\in\mathbb{N}$. Let $u$ be
  an integer such that $u$ is not divisible by $p$. Then, $\gcd\left(
u,p^{\alpha}\right)  =1$.

Proof of Lemma 4. The integer $\gcd\left(  u,p^{\alpha}\right)  $ is a
positive divisor of $p^{\alpha}$, and therefore a power of $p$ (by Lemma 3).
In other words, $\gcd\left(  u,p^{\alpha}\right)  =p^{\beta}$ for some
$\beta\in\mathbb{N}$. Consider this $\beta$. If $\beta>0$, then $p\mid
p^{\beta}=\gcd\left(  u,p^{\alpha}\right)  \mid u$, which contradicts the
assumption that $u$ is not divisible by $p$. Hence, we cannot have $\beta>0$,
and thus we must have $\beta=0$. Hence, $p^{\beta}=p^{0}=1$ and $\gcd\left(
u,p^{\alpha}\right)  =p^{\beta}=1$. This proves Lemma 4. $\blacksquare$

Theorem 5. Let $p$ be a prime number. Let $\alpha\in\mathbb{N}$ and let
  $k$ be an integer such that $0<k<p^{\alpha}$. Then, $\dbinom{p^{\alpha}}{k}$
  is divisible by $p$.

Your claim follows from Theorem 5 (applied to $p=2$ and $\alpha=n$), since your $k$ satisfies $0 < k < n \leq 2^n$.
Proof of Theorem 5. Assume the contrary. Thus, $\dbinom{p^{\alpha}}{k}$ is
not divisible by $p$. Hence, Lemma 4 (applied to $u=\dbinom{p^{\alpha}}{k}$)
that $\gcd\left(  \dbinom{p^{\alpha}}{k},p^{\alpha}\right)  =1$.
Applying Lemma 1 to $n=p^{\alpha}$ and $m=k$, we obtain $\dbinom{p^{\alpha}
}{k}=\dfrac{p^{\alpha}}{k}\cdot\dbinom{p^{\alpha}-1}{k-1}$, so that
$k\dbinom{p^{\alpha}}{k}=p^{\alpha}\dbinom{p^{\alpha}-1}{k-1}$. Thus,
$p^{\alpha}\mid k\dbinom{p^{\alpha}}{k}=\dbinom{p^{\alpha}}{k}k$. Hence, Lemma
2 (applied to $x=p^{\alpha}$, $y=\dbinom{p^{\alpha}}{k}$ and $z=k$) yields
$p^{\alpha}\mid k$ (since $\gcd\left(  p^{\alpha},\dbinom{p^{\alpha}}
{k}\right)  =\gcd\left(  \dbinom{p^{\alpha}}{k},p^{\alpha}\right)  =1$). Since
$k$ is positive, this yields $k\geq p^{\alpha}$; but this contradicts
$k<p^{\alpha}$. This contradiction shows that our assumption was wrong. Thus,
Theorem 5 is proven. $\blacksquare$
