Remark: a) If $n=2k$, the first problem is trivial, because if we denote ${{2}^{n}}-1=m>n$, we have $f(x)={{x}^{m-1}}+{{x}^{m-2}}+...+x+1-{{x}^{2k}}$, so on the basis of Bezout theoreme because we have $f(-1)=0$, result that we have $f$ divisible by $x+1$ And if $n=4k+1$, we have $f(i)=f(-i)=0$, so $f$ is divisible by $x^2+1$. It remain to see the situation $n=4k+3$.

$f_n (x)=\frac {X^{2^n-1}-1}{X-1}-X^n $ a)$1+x^{v_2(n)} | f_n(x)$ b)$\frac{x^7-1}{x-1} | f_7(x)$

rmtf1111 wrote: $f_n (x)=\frac {X^{2^n-1}-1}{X-1}-X^n $ a)$1+x^{v_2(n)} | f_n(x)$ It's not true when n is odd. You can't say that $2|f_n(x)$ for odd n.

$(a)$ $\bullet$ If $n$ is even, choose $p=1$. $\bullet$ If $n$ is odd, choose $p=2^{v_2(n+1)}$

For $(b)$ from $(a)$ we get $x^8+1$ is one of the factors. Can someone find $2$ more factors???