A question about product representation of quantum Fourier transform 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.
 A: 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
$$
\begin{array}{lll}
&& \left[|0\rangle+e^{2\pi i(\frac{j_n}{2})}|1\rangle\right] \\
&+& \left[|0\rangle+e^{2\pi i(\frac{j_{n-1}}{2}+\frac{j_n}{2^2})}|1\rangle\right] \\
&+& \cdots \\
&+& \left[|0\rangle+e^{2\pi i(\frac{j_1}{2}+\frac{j_2}{2^2}+\cdots+\frac{j_n}{2^n})}|1\rangle\right] \\
&=& \left[|0\rangle+e^{2\pi i 0.j_n}|1\rangle\right] \\
&+& \left[|0\rangle+e^{2\pi i 0.j_{n-1} j_n}|1\rangle\right] \\
&+& \cdots \\
&+& \left[|0\rangle+e^{2\pi i 0.j_1 j_2 \cdots j_n}|1\rangle\right]
\end{array}
$$
