 Solution

Contract the numerator of the quoted light cone gauge propagator with $n_{\mu }$:Thus every internal gluon emerging from a ghost-gluon vertex contributes a factor $n_{\mu }{\tilde{\Delta }}^{\mu \nu }=0$. Every ghost-dependent contribution to a physical gluon or matter amplitude vanishes. Equivalently, the strict condition $n\cdot A=0$ removes the Faddeev-Popov ghost interaction in i, leaving the field-independent determinant $\det (n\cdot \partial )$, which cancels from normalized correlators.

This is ghost decoupling in strict axial gauge, not a statement that the free Faddeev-Popov ghost propagator itself is zero. The algebra assumes a consistent axial-gauge pole prescription for $n\cdot p=0$ and treatment of residual gauge zero modes. The overall sign assigned to the displayed vector propagator does not enter the contraction proof. With the mostly-minus convention and the usual $-F^2/4$ action, its conventional prefactor is $-i/(p^2+i0)$; the source prints $+i$. Either prefactor gives the same zero contraction, so no sign correction is needed to obtain the requested ghost-decoupling result.