By induction; case n=1 is true.
$A$ admits an eigenvalue $\lambda$ with eigenvector $v$ over $\mathbb{C}$. Change $A$ into a basis $e_1=v,...,e_n$. Then $\exists X$ such that $XAX^{-1}=\left( \begin{array}{ccc} \lambda & * \\ 0 & B \\ \end{array} \right)$ where $C_A(x)=(x-\lambda)C_B(x)$ $[1]$
$\Rightarrow A=X^{-1}\left( \begin{array}{ccc}\lambda & * \\ 0 & B \\ \end{array} \right)X$ so $C_A(A)=C_A\left[X^{-1}\left( \begin{array}{ccc}\lambda & * \\ 0 & B \\ \end{array} \right)X \right]=X^{-1}C_A\left[\left( \begin{array}{ccc}\lambda & * \\ 0 & B \\ \end{array} \right)\right]X$
Then using $[1]$ we get: $C_A(A)=X^{-1}\left[\left( \begin{array}{ccc}\lambda-\lambda & * \\ 0 & B-\lambda \\ \end{array} \right)\left( \begin{array}{ccc}C_B(\lambda) & * \\ 0 & C_B(B) \\ \end{array} \right)\right]X$ then by the inductive step we have $C_B(B)=0$ so we achieve $C_A(A)=X^{-1}\left[ \left( \begin{array}{ccc}0 & * \\ 0 & B-\lambda \\ \end{array} \right)\left( \begin{array}{ccc}C_B(\lambda) & * \\ 0 & 0 \\ \end{array} \right)\right]X=0$
Thanks.
This seems to be correct. You assume that the field is $\mathbb C$. But when the field is any algebraically closed field, there is an even easier argument using essentially the same induction: first prove using induction starting with the existence of an eigenvalue that the matrix is similar to a triangular matrix; then note that Cayley-Hamilton is obvious for triangular matrices.
This special case implies the general case by abstract nonsense.