Electromagnetic current operator using Feynman rules

Recall where this $Gamma^mu$ comes from. If you look at the top of page 185 in Peskin, he introduces it using the diagram

$phantom{qquad qquadqquad }$

The $Gamma^mu$ is just the "blob" (i.e. effective vertex), meaning that it represents everything that happens in between an electron entering and leaving the vertex (and interacting with some external $A_mu$). Actually, we name the vertex $-ieGamma^mu$ to make the factors of $e$ more clean to deal with.

If we read off the Feynman rules as usual, then, we arrive at the contribution to the matrix element given by

$$bar{u}(p')(-ie Gamma^mu) u(p)$$

But what is $Gamma^mu$ equal to? How would we find it? Well like we said, we know that it is the sum of everything that can happen within the "blob". In the language of Feynman diagrams this means that it is the sum of all diagrams that you can make by inserting only a photon propagator. The reason that we are only allowed to insert photon propagators is nothing deep, it's just that we have simply restricted ourselves to looking only at these possible corrections.

Therefore, we at least know that $Gamma^mu$ is given by

$$ -ieGamma^mu = -iegamma^mu + mathcal{O}(e^3) $$

where the order $e^3$ term is found by connecting the incoming and outgoing lines with a photon propagator (i.e. the "vertex correction"; also, if we divide out the factor of $e$ on both sides then the correction becomes $mathcal{O}(e^2)$, which it is convention to do).

So to answer your question directly, we can use Feynman diagrams because it is essentially built into the definition of $Gamma^mu$, as the sum of all possible photon propagator insertions to the tree level diagram.