number theory - For what $n$ is it true that $(1+sum_{k=0}^{infty}x^{2^k})^n+(sum_{k=0}^{infty}x^{2^k})^nequiv1mod2$

Let $A:=\sum_{k=0}^{\infty}x^{2^k}$. For what $n$ is it true that

$(A+1)^n+A^n\equiv1\mod2$ (here we are basically working in $\mathbb{F}_2$.)

The answer is all powers of 2, and it's fairly simple to see why they work, but the hard part is proving all non-powers of 2 don't work. In the solution, it says

"If $n$ is not a power of 2, say $n=2^i(2j+1),j\ge1$, then the smallest m for which $\binom{n}{m}$ is odd is $2^j$."

I don't see why. I know the highest power of $2$ in $\binom{n}{m}$ is $\sum_{k\ge1}[n/2^k]-[m/2^k]-[(n-m)/2^k]$ where $[.]$ is the greatest integer function. Also each summand is 0 or 1. It's 1 when the sum of fractional parts $\{m/2^k\} + \{(n-m)/2^k\}\ge1$, and 0 otherwise.