Where is the BRST symmetry?

When quantizing YM we start from the gauge fixed path integral (to remove redundancy of integrating over Gauge symmetric configurations) $$\int \mathcal{D}A \delta(G(A)) \text{det} \Delta_{FP}e^{i\int d^4x \mathcal{L}_{YM}} \tag{1}$$ Now, one would rewrite the delta and introduce ghosts to the theory by raising the resulting term into the exponential. Which would then be BRST symmetric. I assume BEFORE we did this, the BRST symmetry was hidden in the Gauge symmetry of the Lagrangian.

But where is the BRST symmetry in the above gauge fixed integral? We haven't introduced any ghosts particles yet. Adding a source term would change the above integral into a generating functional defining our whole theory WITHOUT any ghosts appearing.

The BRST symmetry cannot be seen without introducing auxiliary variables. The fastest way to realize the BRST symmetry is to "exponentiate" the delta function $$\delta(G)~=~\int \!{\cal D} B ~\exp\left[iB_{\alpha}G^{\alpha}\right]$$ and the Faddeev-Popov (FP) determinant $$\det\Delta ~=~\int \!{\cal D} c ~{\cal D} \bar{c} ~\exp\left[\bar{c}_{\alpha} \Delta^{\alpha}{}_{\beta} ~c^{\beta}\right]$$ by introducing Lagrange multipliers $B_{\alpha}$ and FP Grassmann-odd ghosts $c^{\alpha}$ and antighosts $\bar{c}_{\alpha}$ in the action, see. e.g. this Phys.SE post (for abelian gauge group).
And voila: the extended action (with the auxiliary variables $B$, $c$, $\bar{c}$) has BRST symmetry!