# Superpotential is Supergauge Invariant?

I was studying by the X. Tata & Bauer and I'm stuck with something: In Chapter 6 - Supersymmetric Gauge Theories, it states that the superpotential is already invariant by a supergauge transformation, and explains that is because "it's polynomial in the chiral superfields".

I'm thinking about it, but I guess i'm too tired or something, but I can't prove that every polynomial function of the chiral superfields will be supergauge invariant....

Could someone give me a tip?

A chiral superfield with variable change $x_\mu\rightarrow x_\mu+ \frac{1}{2}\bar\theta \gamma_5\gamma_{\mu}\theta$ will have similar gauge transformation as it's components and if the action does not contain derivative or complex conjugate of the chiral superfield and contains a polynomial part then under a local gauge transformation involving $exp(it_A\Omega_A(x))$ action will be invariant if it is invariant under global transformation with $\Omega$ independent of $x_\mu$.