Show that the square matrix A is invertible

If $p(0)$ is nonzero then $a_0$ is nonzero. Therefore, one has: $$I=-\sum_{i=1}^n\frac{a_i}{a_0}A^i=-A\sum_{i=0}^{n-1}\frac{a_{i+1}}{a_0}A^i.$$

There's a simple way to do this without manipulating an expansion of the polynomial, or knowing anything about determinants or characteristic polynomials.

First, recall that $A$ is non-invertible if and only if there exists some non-zero vector $\bf{x}$ such that $A{\bf x} = {\bf 0}$. Suppose that there exists such a vector.

Since $p(A)$ is a sum of matrices ($a_n A^n + \dots + a_1 A + a_0 I$) we may compute ${\bf x}^T p(A) \bf{x}$. The fact that $A {\bf x} = 0$ implies that ${\bf x}^T A^n {\bf x} = {\bf 0}$ for any $n>0$, so we have that ${\bf x}^T p(A) {\bf x} = {\bf x}^T a_0 {\bf x} = p(0) \left| {\bf x} \right|^2$. Since we are told that $p(A) = 0$, we have that $p(0) \left| {\bf x} \right|^2 = 0$. Since ${\bf x}$ is non-zero, we must have $p(0) = 0$, contradicting the assumption that $p(0) \ne 0$.

To summarise: we are given that $p(A) = 0$ and that $p(0) \ne 0$. If $A$ is not an invertible matrix, then the argument of the previous two paragraphs shows that the conditions $p(A) = 0$ and $p(0) \ne 0$ cannot both hold. So $A$ must be invertible.

$p(0)\neq 0$ implies that $0$ is not a root of characteristic polynomial $p(x)$ which is turn says that $0$ is not an eigenvalue of $A$. As $det(A)=$ product of eigenvalues gives $det(A)\neq 0$ which suggests that $A$ is invertible.