How do I show that
$$ \phi(x)(xI-A)^{-1}= M_0x^{n-1}+ M_1x^{n-2} + \cdots + M_{n-2}x + M_{n-1}$$
where $A$ is a $n$ dimension square matrix, I is an identity matrix, $\phi(x)=x^n + a_1 x^{n-1} \cdots a_{n-1}x + a_n$ is the characterstic polynomial of $A$ and $M_k$ is a $n$ dimension square matrix in which $k=0,1, \ldots , n-1 $.
I started by first expressing the inverse: $$(xI-A)^{-1}= \det(xI-A)^{-1} \operatorname{adj}(xI-A)$$
Since $\phi(x)$ is the charateristic polynomial of $A$, $$\phi(x)=\det(xI-A)$$ Hence, $$\phi(x)(xI-A)^{-1}= \operatorname{adj}(xI-A)$$
Is what I have done up to now, correct? If so, what would be my next step?
I find no problem in your first step. Then the next step should be to describe $adj(xI-A)$. Let's denote $adj(xI-A) = (c_{ij})$
Note that $c_{ij}$ is the determinant of the matrix $xI-A$ deprived of the i-th column and the j-th row, up to multiplied by $-1$. So $c_{ij}$ is a polynomial with degree at most equal to $n-1$. Let's write $c_{ij} = \sum_{k=0}^{n-1} p^{ij}_kx^{n-1-k} $.
Then define $M_k = (p_{k}^{ij})_{i,j}$, we get the conclusion