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.