Characteristic polynomial of a matrix is monic?

You have $$c_A(\lambda) = \det(\lambda I - A) = \sum_{\sigma\in \Sigma_n} {\rm sgn}(\sigma) \prod_{i=1}^n (\lambda \delta_{i\sigma(i)} - A_{i\sigma(i)}),$$ where $\Sigma_n$ is the permutation group on $n$ letters. The only term in the sum giving a $\lambda^n$ term is the identity permutation. Hence the polynomial is monic.

$$A=\begin{pmatrix}a_{11}&a_{12}&\ldots&a_{1n}\\a_{21}&a_{22}&\ldots&a_{2n}\\\ldots&\ldots&\ldots&\ldots\\a_{n1}&a_{n2}&\ldots&a_{nn}\end{pmatrix}\implies $$

$$|xI-A|=\begin{vmatrix}x-a_{11}&-a_{12}&\ldots&-a_{1n}\\-a_{21}&x-a_{22}&\ldots&-a_{2n}\\\ldots&\ldots&\ldots&\ldots\\-a_{n1}&-a_{n2}&\ldots&x-a_{nn}\end{vmatrix}=\prod_{i=1}^n(x-a_{ii})+\ldots=x^n+\ldots$$

and this is enough since we know $\;\deg p_A(x)=n\;$ ...

If you already know that $P_A$ is a polynomial, and if the matrix has real or complex coefficients, here is a nice trick: $$ \lim_{|x|\to+\infty} \frac{P_A(x)}{x^n} = \lim_{|x|\to+\infty} \det \left( I - \frac{1}{x} A \right) = \det I = 1, $$ because $\det$ is continuous. From this, you can deduce that $P_A$ is monic of degree $n$.