Combinatorial proofs that $\binom{2^n}{k}$ is even for every $n\geq 1$ and $1\leq k \leq 2^n-1$. I am trying to prove that $\binom{2^n}{k}$ is even for every $n\geq 1$ and $1\leq k \leq 2^n-1$.
The one I could come up with is that the equivalence relation where $A\sim B$ if $B$ is of the form $A+k$ for some $k$ with the addition taken $\bmod 2^n$ splits the subsets into classes, no class has size $1$ and the size is a divisor of $2^n$ so they are all even. I am wondering if there are other combinatorial proofs.


I am trying to prove hat $\binom{2^n}{k}$ is even for every $n\geq 1$ and $1\leq k \leq 2^n-1$.

Assumed that it is intended that $n,k \in \Bbb{Z}.$
Alternative (direct) approach that uses
Legendre's Formula.
While less elegant than other approaches, this approach provides a nice fail-safe, if elegance eludes you.

For $p$ prime, and $n \in \Bbb{Z_{\geq 2}},~$ let $~V_p(n)~$ denote the largest non-negative integer $~\alpha~$ such that $~p^\alpha ~| ~(n!)$.
For $r \in \Bbb{R}$, let $\lfloor r\rfloor$ denote the largest integer $k$ such that $k \leq r.$ 
That is, $\lfloor r\rfloor$ is the floor of $r$.

Preliminary result
(1)
Given: $~\displaystyle a,b,c \in \Bbb{Z^+}, 
~\left\lfloor \frac{a}{c} \right\rfloor ~+
~\left\lfloor \frac{b}{c} \right\rfloor ~\leq
~\left\lfloor \frac{a + b}{c} \right\rfloor.$ 
Proof:
Denote $\displaystyle ~\left(\frac{a}{c}\right) ~\text{as}
 ~P_1 + r_1, ~~~\text{and} ~~~
~\left(\frac{b}{c}\right) ~\text{as}
 ~P_2 + r_2$ 
where $~P_1, P_2 \in \Bbb{Z}, ~0 \leq r_1, r_2 < 1.$ 
Then, $~\displaystyle 
~\left\lfloor \frac{a}{c} \right\rfloor ~+
~\left\lfloor \frac{b}{c} \right\rfloor = P_1 + P_2.$
$\underline{\text{Case 1:} ~~0 \leq (r_1 + r_2) < 1}.$
Then $\displaystyle ~\left\lfloor \frac{a + b}{c} \right\rfloor = P_1 + P_2.$
$\underline{\text{Case 2:} ~~1 \leq (r_1 + r_2) < 2}.$
Then $\displaystyle ~\left\lfloor \frac{a + b}{c} \right\rfloor = P_1 + P_2 + 1.$

Per Legendre's formula, 
$\displaystyle V_p(n) = \sum_{i=1}^\infty \left\lfloor \frac{n}{p^i}\right\rfloor. $ 
It is understood that the start of the summation consists of a finite number of non-zero terms, and that all of the remaining terms in the summation are $= (0)$.
For $\displaystyle 1 \leq k \leq 2^n - 1, ~\binom{2^n}{k} = \frac{\left(2^n\right)!}{k!\left[\left(2^n\right) - k\right]!}.$
$$\displaystyle V_2\left(2^n\right) = \sum_{i=1}^{n-1} \left\lfloor \frac{2^n}{2^i}\right\rfloor
 ~~+ ~~\left\lfloor \frac{2^n}{2^n}\right\rfloor. \tag{A}$$
$$\displaystyle V_2\left(k\right) = \sum_{i=1}^{n-1} \left\lfloor \frac{k}{2^i}\right\rfloor
 ~~+ ~~\left\lfloor \frac{k}{2^n}\right\rfloor. \tag{B}$$
$$\displaystyle V_2\left(2^n - k\right) = \sum_{i=1}^{n-1} \left\lfloor \frac{2^n - k}{2^i}\right\rfloor
 ~~+ ~~\left\lfloor \frac{2^n - k}{2^n}\right\rfloor. \tag{C}$$
Since $k < 2^n~~~$ and $~~~\left(2^n - k\right) < 2^n$ 
the right hand terms in (B) and (C) above are both $= 0$, 
while the right hand term in (A) $= 1.$
For each of the other $(n-1)$ terms in the summations of (A), (B), and (C) 
you can invoke Preliminary Result (1) above.
That is:
$\displaystyle \left\lfloor \frac{2^n}{2^1} \right\rfloor
~\geq ~\left\lfloor \frac{k}{2^1} \right\rfloor
~+ ~\left\lfloor \frac{2^n - k}{2^1} \right\rfloor.$
$\displaystyle \left\lfloor \frac{2^n}{2^2} \right\rfloor
~\geq ~\left\lfloor \frac{k}{2^2} \right\rfloor
~+ ~\left\lfloor \frac{2^n - k}{2^2} \right\rfloor.$
$\cdots$
$\displaystyle \left\lfloor \frac{2^n}{2^{(n-1)}} \right\rfloor
~\geq ~\left\lfloor \frac{k}{2^{(n-1)}} \right\rfloor
~+ ~\left\lfloor \frac{2^n - k}{2^{(n-1)}} \right\rfloor.$
Therefore, the summation of $(n-1)$ terms in (A) is $\geq$ the corresponding combined summations in (B) and (C) above.
Therefore,
$$  V_2\left(2^n\right) > V_2\left(k\right) + 
V_2\left(2^n - k\right).$$

Addendum
As a corrollary to this answer's analysis, you can immediately conclude that
$$2^1 ~| ~\binom{2^n}{2^{[n-1]}}~~~\text{and}
~~~2^2 ~\not| ~\binom{2^n}{2^{[n-1]}}.$$
This follows, per examination of (A), (B), and (C) in the answer by noting that when $~c|a~$ and $~c|b~$ then
$$\left\lfloor \frac{a}{c} \right\rfloor + 
\left\lfloor \frac{b}{c} \right\rfloor = 
\left(\frac{a}{c} + \frac{b}{c}\right) = 
\left(\frac{a + b}{c}\right) = 
\left\lfloor \frac{a + b}{c} \right\rfloor.$$
