$A ∈ GL_n(K)$. Show that there exists a polynomial $P(X) ∈ K[X]$ of degree $< n$ such that $A^{−1} = P(A)$

Question

Hey I am having problems with this exercise. Can someone help me?

Let $A ∈ GL_n(K)$. Show that there exists a polynomial $P(X) ∈ K[X]$ of degree $< n$ such that $A^{−1} = P(A)$.

To show that there exists a polynomial $P(X) \in K[X]$ of degree $< n$ such that $A^{−1} = P(A)$, I have used of the fact that the characteristic polynomial of $A$ is given by $p(x) = \det(xI - A)$.

Since $A$ is invertible, its characteristic polynomial $p(x)$ has the form $p(x) = x^n + c_{n-1}x^{n-1} + \ldots + c_1x + c_0$, where $c_0 = \det(-A) \neq 0$.

Now, let us consider the polynomial $q(x) = x^{n-1} + c_{n-1}x^{n-2} + \ldots + c_1$. Notice that $q(A)$ has degree $< n$, since the highest degree term of $q(A)$ is $A^{n-1}$.

We claim that $A^{-1} = q(A)$. To see this, we will show that $q(A)A = Aq(A) = I$.

First, let us show that $Aq(A) = I$. We have:

\begin{align*} Aq(A) &= A(A^{n-1} + c_{n-1}A^{n-2} + \ldots + c_1) \\ &= A^n + c_{n-1}A^{n-1} + \ldots + c_1A \\ &= p(A)-c_0=c_0 \neq I \end{align*}

And here I am having my problem, how can I continue?

Does what I've done up until now make any sense?