${2^k-1\choose a}$ is odd?

Question

How can I see that $${2^k-1\choose a}\equiv 1\mod 2$$ for $0<a<2^k$?

I tried playing around with ${2^k-1\choose a}=\frac{(2^k-1)!}{a!(2^k-1-a)!}$ but it didn't get me very far.

Answer 1

Show that $\binom{2^k}{a}$ is even for $0<a<2^k$ which is equivalent to prove $$(1+x)^{2^k}\equiv 1+x^{2^k}\pmod{2}.$$ This can be seen by induction by noting that $$(1+x)^{2^{k}}=(1+x)^{2^{k-1}}\cdot (1+x)^{2^{k-1}} \equiv (1+x^{2^{k-1}})(1+x^{2^{k-1}})\\=1+2x^{2^{k-1}}+x^{2^{k}}\equiv 1+x^{2^k}\pmod{2}.$$

Answer 2

You are looking at the coefficient of $x^{a}$ in the expansion of $$ (1 + x)^{2^{k} - 1} $$ modulo $2$.

Now modulo $2$ one has $$ (1 + x) (1 + x)^{2^{k} - 1}= (1 + x)^{2^{k}} = 1 + x^{2^{k}} = (1 + x)(1 + x + \dots + x^{2^{k} - 1}). $$