Faddeev-Popov ghosts are introduced in the quantization of Yang-Mills theory to absorb the Faddeev-Popov determinant into the action, $$\det \Delta_{\text{FP}} = \int \mathcal{D} \bar{c} \mathcal{D} c \, e^{i \int dx \, \bar{c}^a(x) (\Delta_{\text{FP}} c(x))^a}.$$ Here, $c$ is a Lorentz scalar, and it must be fermionic, as if we used a bosonic variable $\psi$ instead, we would get the determinant in the denominator rather than the numerator, $$\frac{1}{\det \Delta_{\text{FP}}} \propto \int \mathcal{D} \bar{\psi} \mathcal{D} \psi \, e^{i \int dx \, \bar{\psi}^a(x) (\Delta_{\text{FP}} \psi(x))^a}.$$ So we have to accept a violation of spin-statistics to quantize Yang-Mills.
But a friend of mine came up with a simple alternative: simply note that $$\det \Delta_{\text{FP}} = \frac{1}{\det \Delta_{\text{FP}}^{-1}} \propto \int \mathcal{D} \bar{\psi} \mathcal{D} \psi \, e^{i \int dx \, \bar{\psi}^a(x) (\Delta_{\text{FP}}^{-1} \psi(x))^a}.$$ Note that the inverse exists because the determinant is nonzero; if the determinant were zero the whole path integral would be zero and we wouldn't be able to do anything, ghosts or not.
Is there anything wrong with this method? Does it lead to some complications down the line? If not, why are the spin-statistics violating Faddeev-Popov ghosts typically used instead, when this setup looks much nicer?