Why is n a power of two for n over k even

Question

If $\binom{n}{k}$ is even for every $1 \le k \lt n$, $n$ has to be a power of two. I'm searching for an elementary way to proof this.

Answer 1

Otherwise write $n=2^rm$ with $m\ge3$ odd, and consider $$\binom{n}{2^r}=\frac{n(n-1)(n-2)\cdots (n-2^r+1)}{2^r(2^r-1)(2^r-2)\cdots(2^r-2^r+1)}.$$ Shot that for $0\le t<2^r$, the same power of two divides $n-t$ and $2^r-t$,