Let $$A(L) = I_K - A_1L - A_2L^2 -\dots - A_pL^p$$ be polynomial of order $p$, where $A_i$ is $K \times K$ matrces for $i=1, \dots, p$ and $L$ is a scalar.
Denote $\gamma(L) = \left[\det A(L)\right]^{-1}$. Then, $\gamma(L)$ can be represented as an infinite order polynomial:
$$\gamma(L) = 1 + \gamma_1L + \gamma_2L^2 + \dots$$
I am looking for an explanation of why $\gamma(L)$ can be indeed represented like this.
I can't cope with $L$ as the polynomial variable so I am going to call it $X$. Nor with $K$ as the dimension so I am going to call it $n$.
Each of the $n^2$ elements of the matrix $A(X)$ is a polynomial of degree at most $p$, and so $\det A(X)$ is a polynomial of degree at most $np$. Moreover, by substituting $X=0$ we see that the constant term of $\det A(X)$ is $1$. That is $\det A(X)=1-Xg(X)$ for some polynomial $g$.
The question is now whether the inverse of any such $1-Xg(X)$ can be expressed $$(1-Xg(X))^{-1}=1+\sum_{1}^{\infty} a_j X^j.$$
Note well, this is not a polynomial. There are no such things as polynomials of infinite degree.
But we can easily find such an expression. By the usual Binomial Theorem, if for each $N$ we ignore terms of degree greater than $N$, then $$ 1+\sum_{1}^{N} a_j X^j=1+ \sum_{1}^{N} X^j g(X)^j,$$ which lets us calculate the coefficients $a_j$ from those of $g$ and so, ultimately, from the entries of your $A_j$.
All this makes sense if we work with formal power series (a proper name for your "infinite" polynomials). But whether it makes sense when we want to do our sums in the usual real or complex numbers depends on the specifics of your $A_j$; these sums may converge; or again, they may not.