We proceed by induction on $k$. For $k=1$ the statement is trivial. Assuming we have proved it for $k=j-1$, we now prove it for $k=j$. Case 1. $n=2 t-1$ for some positive integer $t$. Observe that $$1+\frac{2^{j}-1}{2 t-1}=\frac{2\left(t+2^{j-1}-1\right)}{2 t} \cdot \frac{2 t}{2 t-1}=\left(1+\frac{2^{j-1}-1}{t}\right)\left(1+\frac{1}{2 t-1}\right) .$$ By the induction hypothesis we can find $m_{1}, \ldots, m_{j-1}$ such that $$1+\frac{2^{j-1}-1}{t}=\left(1+\frac{1}{m_{1}}\right)\left(1+\frac{1}{m_{2}}\right) \cdots\left(1+\frac{1}{m_{j-1}}\right)$$ so setting $m_{j}=2 t-1$ gives the desired expression. Case 2. $n=2 t$ for some positive integer $t$. Now we have $$1+\frac{2^{j}-1}{2 t}=\frac{2 t+2^{j}-1}{2 t+2^{j}-2} \cdot \frac{2 t+2^{j}-2}{2 t}=\left(1+\frac{1}{2 t+2^{j}-2}\right)\left(1+\frac{2^{j-1}-1}{t}\right)$$ noting that $2 t+2^{j}-2>0$. Again, we use that $$1+\frac{2^{j-1}-1}{t}=\left(1+\frac{1}{m_{1}}\right)\left(1+\frac{1}{m_{2}}\right) \cdots\left(1+\frac{1}{m_{j-1}}\right) .$$ Setting $m_{j}=2 t+2^{j}-2$ then gives the desired expression.