Inverse of a matrix is expressible as a polynomial?

Let $A$ be an $n \times n$ matrix. Prove that if A is invertible, then there exists a polynomial $p$, such that $A^{-1}=p(A)$

Thus far:

Let $W$ denote the $k$ dimensional A-cyclic subspace spanned by a vector $v$.

Then, $I_n=\sum_{i=0}^{k} a_{i}A^i$ for some scalar $a_i$.

What I want to do is multiply both sides by $A^{-1}$, so that way it just falls into my hands, but Im not sure that is 'legal'.

Hint: The Cayley-Hamilton theorem tells you that $p(A)=0_{n\times n}$ for some polynomial $p(x)$, where $p(x)=\sum \limits_{k=0}^n\left(a_kx^n\right)$.

Since $a_0=(-1)^n\det(A)\neq 0$ you can get something like what you got in your question, only without the zero power problem.