The square modulus of an amplitude must be real. Given that, I am having some trouble understanding the square modulus of a path integral being absolutely real. Given \begin{equation} \int\!Dq(t)\equiv \lim\limits_{N\to\infty}\left( \frac{m}{2\pi i\hbar \delta t} \right)^{\!\frac{N}{2}}\left( \prod^{N-1}_{k=1} \int dq_k \right) , \end{equation}
and \begin{equation} \big\langle \psi_2 \big| e^{-i\hat H T/\hbar}\big| \psi_1 \big\rangle=\int\!Dq(t)\,\exp\left\{\frac{i}{\hbar}\int_0^t\!dt'\,\frac{1}{2}m \dot q_k^2\right\}, \end{equation}
introduce concise notation such that \begin{equation}\label{eq:od7diyh} \int\!Dq(t)\equiv \lim\limits_{N\to\infty}\left( i\beta\right)^{\!\frac{N}{2}}\big[ F(q) \big] ,\quad\text{and}\quad\big\langle \psi_2 \big| e^{-i\hat H T/\hbar}\big| \psi_1 \big\rangle=\int\!Dq(t)\,e^{i\gamma}. \end{equation}
Quantum mechanics requires that \begin{equation} \big\|\big\langle \psi_2 \big| e^{-i\hat H T/\hbar}\big| \psi_1 \big\rangle\big\|^2\in\mathbb{R}~~,\nonumber \end{equation}
so it follows that we must have \begin{equation} \left\{\lim\limits_{N\to\infty}\left( i\beta\right)^{\!\frac{N}{2}}\big[ F(q) \big]e^{i\gamma}\right\}\left\{\lim\limits_{N\to\infty}\left( -i\beta\right)^{\!\frac{N}{2}}\big[ F(q) \big]e^{-i\gamma}\right\}\in\mathbb{R}~~.\nonumber \end{equation}
This simplifies as \begin{equation} \lim\limits_{N\to\infty}\beta^N\big[ F(q) \big]^2 e^{i\gamma}e^{-i\gamma}~~.\nonumber \end{equation}
If I cancel the two exponentials, that will mean that the probability (squared amplitude) doesn't depend on the action at all. If I don't cancel them (for some reason that is not clear to me), then how can I guarantee that the expression is real? What is going on here?