In the Nielsen and Chuang's Quantum Computing and Quantum Information, the last step of proving the product representation of quantum Fourier transform is $$ \frac{1}{2^{n/2}}\bigotimes_{l=1}^{n} \left[ |0\rangle+e^{2\pi i j 2^{-l}}|1\rangle \right] =\frac{ \left(|0\rangle+e^{2\pi i 0.j_n}|1\rangle\right) \left(|0\rangle+e^{2\pi i 0.j_{n-1}j_n}|1\rangle\right)\cdots \left(|0\rangle+e^{2\pi i 0.j_1 j_2\cdots j_n}|1\rangle\right)}{2^{n/2}}. $$
I think there is a mistake. For example, if $j=3=11_{2}$ and $l=1$, then $$ |0\rangle+e^{2\pi i \cdot 3 \cdot 2^{-1}}|1\rangle \neq |0\rangle+e^{2\pi i \cdot 0.1_{2}}|1\rangle $$ Do I misunderstand anything?
The same question has been asked on overflow, but it doesn't attract much attention and the answer is unclear.
The key point is $e^{2\pi i m}=1$ if $m$ is a positive integer. $$ \begin{array}{lll} \bigotimes_{l=1}^{n} \left[|0\rangle+e^{2\pi ij2^{-l}}|1\rangle\right] &=& \bigotimes_{l=1}^{n} \left[|0\rangle+e^{2\pi i\frac{j_1 2^{n-1}+j_2 2^{n-2}+\cdots +j_n 2^0}{2^{l}}}|1\rangle\right] \\ &=& \left[|0\rangle+e^{2\pi i\frac{j_1 2^{n-1}+j_2 2^{n-2}+\cdots +j_n 2^0}{2}}|1\rangle\right] \\ &+& \left[|0\rangle+e^{2\pi i\frac{j_1 2^{n-1}+j_2 2^{n-2}+\cdots +j_n 2^0}{2^2}}|1\rangle\right] \\ &+& \cdots \\ &+& \left[|0\rangle+e^{2\pi i\frac{j_1 2^{n-1}+j_2 2^{n-2}+\cdots +j_n 2^0}{2^n}}|1\rangle\right] \\ &=& \left[|0\rangle+e^{2\pi i(m_1+\frac{j_n}{2})}|1\rangle\right] \\ &+& \left[|0\rangle+e^{2\pi i(m_2+\frac{j_{n-1}}{2}+\frac{j_n}{2^2})}|1\rangle\right] \\ &+& \cdots \\ &+& \left[|0\rangle+e^{2\pi i(\frac{j_1}{2}+\frac{j_2}{2^2}+\cdots+\frac{j_n}{2^n})}|1\rangle\right] \end{array} $$
Since $e^{2\pi i m}=1$ if $m$ is a positive integer, we can omit them. So the previous expression is $$ \begin{array}{lll} && \left[|0\rangle+e^{2\pi i(\frac{j_n}{2})}|1\rangle\right] \\ &+& \left[|0\rangle+e^{2\pi i(\frac{j_{n-1}}{2}+\frac{j_n}{2^2})}|1\rangle\right] \\ &+& \cdots \\ &+& \left[|0\rangle+e^{2\pi i(\frac{j_1}{2}+\frac{j_2}{2^2}+\cdots+\frac{j_n}{2^n})}|1\rangle\right] \\ &=& \left[|0\rangle+e^{2\pi i 0.j_n}|1\rangle\right] \\ &+& \left[|0\rangle+e^{2\pi i 0.j_{n-1} j_n}|1\rangle\right] \\ &+& \cdots \\ &+& \left[|0\rangle+e^{2\pi i 0.j_1 j_2 \cdots j_n}|1\rangle\right] \end{array} $$