Because the lhs and the rhs are both multilinear, it is sufficient to prove the "interior product formula" when $\omega$ and $\mu$ are decomposable, i.e. wedge products of $1$-forms.
Working in such a reduced context our formula will be an instance of $\iota_X(\omega_1\wedge\ldots\wedge\omega_p)=\sum_{i=1}^p(-1)^{i-1}(\iota_X\omega_i)\omega_1\wedge \ldots\wedge\widehat{\omega_i}\wedge\ldots\wedge\omega_p.\tag{1}$ (Where $\widehat{\phantom{a}}$ denotes an omitted factor.)
The formula $(1)$ holds if and only if its two sides assume the same value on any $(X_2,\ldots,X_p).$
So, using the definition of interior product and denoting $X=X_1,$ we have to show that $(\omega_1\wedge\ldots\wedge\omega_p)(X_1,\ldots,X_p)=\\=\sum_{i=1}^p(-1)^{i-1}(\iota_X\omega_i)(\omega_1\wedge \ldots\wedge\widehat{\omega_i}\wedge\ldots\wedge\omega_p)(X_2,\ldots,X_p)\tag{2}$ If $A$ is the matrix $(\omega_i(X_j))_{i,j=1,\ldots,p},$ then the lhs of $(2)$ is $\det(A)$ while the rhs of $(2)$ is the Laplace expansion of $\det(A)$ w.r.t. the first column of $A$.