How is the perturbative expansion justified in QED if we make $A_{\mu}\to{}\frac{A_{\mu}}{e}$? Consider the QED Lagrangian
$$\mathcal{L}=\bar{\Psi}(i\gamma^{\mu}D_{\mu}-m)\Psi-\frac{1}{4}F_{\mu\nu}^2$$
where $D_{\mu}\Psi=\partial_{\mu}\Psi-ieQA_{\mu}$ where $e$ is positive. For concreteness' sake consider the case of the electron $Q=-1$. Consider now correlation functions. As it's well known, these satisfy
$$\langle\Omega|T\{\ldots{}e^{i\int{}d^4x\,\mathcal{L}_I}\}|\Omega\rangle=\frac{\langle{}0|T\{\ldots{}e^{i\int{}d^4x\,\mathcal{L}_I}\}|0\rangle}{\langle{}0|T\{e^{i\int{}d^4x\,\mathcal{L}_I}\}|0\rangle}.$$
where for QED we have that $\mathcal{L}_I=-e\bar{\Psi}\gamma^{\mu}A_{\mu}\Psi$. Expanding the exponential we obtain a power series in $e$, which is to my understanding the perturbative expansion of QED. Consider now a little twist. Consider the same Lagrangian with the redefinition
$$A_{\mu}\to{}\frac{A_{\mu}}{e}$$.
This makes the Lagrangian be 
$$\mathcal{L}=\bar{\Psi}(i\gamma^{\mu}D_{\mu}-m)\Psi-\frac{1}{4e^2}F_{\mu\nu}^2$$
where now the interaction part is $\mathcal{L}_I=\bar{\Psi}\gamma^{\mu}A_{\mu}\Psi$, that is without any $e$. Consequently, we can no longer expand in the parameter $e$. What have I forgotten?
 A: The expansion parameter must be (any) parameter you multiply the field by, because it has to be reminiscent of the fact that the field must actually interact with the charge to generate interaction. In this way, should the charge be zero then no interaction would happen, which is exactly the observed behaviour. If you divide by the charge instead, letting $e\to 0$ would generate divergencies which would not make sense; thus that could not be an allowed Lagrangian. One may nevertheless divide by any parameter at will, but there must be a multiplicative coefficient anyway for the reasons explained above. That coefficient will then be promoted to the expansion parameter to expand by (in the case at hand it is actually a combination of $e$ and $\hbar$ rather than $e$ alone).
From the mathematical point of view, the expansion is anyway an expansion in the order of the integrals, namely the first term contains one integral, the second term contains two and so on and so forth; consequently it would not really matter what coefficient you expand by (although eventually you may want to reconstruct the physical transition amplitudes for scattering, therefore reducing back to the above argument).
