Why is $\det(A-\lambda I)=(\lambda-c)^n$ when $(A-cI)^n=0$? Let $A$ be a $n\times n$ matrix and suppose that $(A-cI)^n=0$ for some scalar $c$.
Then why the characteristic polynomial of $A$ is $(x-c)^n$?
 A: We have the following proof by contrapositive:
Suppose that the characteristic polynomial of $A$ has a factor $(x-d)$ with $d \neq c$. Then $A$ has an eigenvector $v$ associated with $d$, and
$$
(A-cI)^n v = (d-c)^n v \ne 0
$$
It follows that $(A-cI)^n \ne 0$.

Argument for algebraically closed fields:
Let $m(x)$ be the minimal polynomial of $A$, and $p$ the characteristic polynomial.  We have the following two statements:


*

*if $q(A) = 0$ for some polynomial $q(x)$, then $m \mid q$ (easy to show)

*if an irreducible polynomial divides $p$ if and only if it divides $q$ (more difficult)


These statements are enough to deduce the characteristic polynomial of you matrix in non-closed fields.
A: Note that if $\lambda$ is an eigenvalue of $A$ then $\lambda-c$ is an eigenvalue of $A-cI$, and hence $(\lambda-c)^n$ is an eigenvalue for $(A-cI)^n=0$.
This proves that $(\lambda-c)^n=0$ and thus $\lambda=c$.
I just proved that $\lambda=c$ is the only eigenvalue of $A$ . This implies what you want.
The above argument works in any algebraically closed field, and if your original field is not algebraically closed, you go to the algebraic closure and make the argument. 
If the field is not algebraically closed, proving that $\lambda=c$ is the only eigenvalue is not enough to conclude that $\det(A-\lambda I)=(c-\lambda)^n$.
P.S. Alternative approach: Let $P(X)$ be the minimal polynomial of $A$. Then $P(X) |(X-c)^n$. 
The Hamilton Cayley Theorem doesn't help in the standard form, but there is a stronger version which sais that $P(X)$ and the characteristic Polynomial of $A$ have the same roots in the algebraic closure of your field.
