If is possible to calculate ${\rm ord}_2((2k+1)^n-1)$ where ${\rm ord}_2(N)$ means the greatest non-negative integer $\nu$ such that $2^\nu$ divides $N$ and $n$ is odd?
After simplification I have $$ (1+2k)^n-1=1+\binom{n}{1} 2k+\binom{n}{2} (2k)^2+(2k)^n-1=2k\left(n+\binom{n}{2} (2k)+\cdots+(2k)^{n-1}\right). $$ Thus $$ {\rm ord}_2((2k+1)^n-1)=1+{\rm ord}_2(k)+{\rm ord}_2(\underbrace{n+\binom{n}{2} (2k)+\cdots+(2k)^{n-1}}_{\text{odd number }})=1+{\rm ord}_2(k)+0. $$ Am I right?