# Clarification about QTF proof regarding equality of QFT application and circuit application

I'm self learning quantum computing through IBM's Qiskit's learning section (which I really like), and I've stumbled across an inequality that I don't quite understand fully.
This must be really easy, I just don't get this next inequality.

In their of the QFT circuit implementation, they write:

What interests me is the last equation, and specifically in the last equation the x circled in blue. This interest me because of their next proof of the QFT's circuit implementation:
Again, the last equation is what interests me.

Disregarding the order of qubits (Which I know is reversed in the equations), I do not understand how the first picture x's are equal to the second picture x's.
For example, in their example for a 3-qubit QFT, their output of the circuit is as follows: Which is to my understanding not equal to their equation in the first picture (Focusing on the x's - in the first picture the x's are complete in each tensor product of the equation while in the last its only x's partial components in each tensor product).

I suppose there are two sources of confusion here -- one is the reversed order of the qubits between the circuit and the unitary, and the other is the use of $$x$$.
In terms of qubit ordering, I think your $$|\psi_6\rangle$$ is the output of the $$\text{QFT}$$ unitary rather than the output of the circuit, i.e. the first and last qubits are swapped.
In terms of the notation, let's consider the three-qubit example. Write $$x = (x_1 x_2 x_3) = x_1 2^{2} + x_{2} 2^1 + x_3$$ in binary, where $$|x_1 x_2 x_3\rangle$$ is the input to the circuit. Then we have for example that $$\frac{x}{2^2} = x_1 + x_2 2^{-1} + x_3 2^{-2}$$ and so the middle term of $$|\psi_6\rangle$$ should be $$\frac{1}{\sqrt{2}} \left[ |0\rangle + \exp\left( 2\pi i \frac{x}{2^2} \right) |1\rangle \right]$$, where $$\exp\left( 2\pi i \frac{x}{2^2} \right) = \exp\left( 2\pi i (x_1 + x_2 2^{-1} + x_3 2^{-2}) \right)$$ $$= \underbrace{\exp\left( 2\pi i x_1 \right)}_{=1} \cdot \exp\left( 2\pi i ( x_2 2^{-1} + x_3 2^{-2}) \right)$$ $$= \exp\left( \frac{2\pi i}{2^2} x_3 + \frac{2\pi i}{2} x_2 \right),$$ which is what we see in the expression.
In other words, we only see the partial components $$x_i$$ of $$x$$ in the last expression because the integer part of $$x / 2^i$$ gets cancelled out in the exponent. Hopefully you can convince yourself that the first and last terms follow similarly?