$D=\begin{bmatrix}0 & 0 & 0 & ... & 0& 0\\n & 0 & 0 & ... & 0 & 0\\0 &n-1 & 0 & ...& 0& 0\\...\\0& 0 & 0 & ... &1&0\end{bmatrix}$
D is the matrix representation of the linear transformation $D : p → \frac{dp}{dx} $ where p is the polynomial of degree n. How can I find the Jordan Canonical Form of D? I now that $|\lambda I - A| = \lambda ^n $, will this help?
I am going to restate the problem before explaining. Let $V$ be the vector space of polynomials of degree less than or equal to $n-1$. I'll assume that the field your were working over is either $\mathbb{R}$ or $\mathbb{C}$, but any field of characteristic zero is fine.
Define $D: V \to V$ by $D(p(x)) = p'(x)$. Note that the $n^{th}$ derivative of any polynomial of degree $(n-1)$ or less is $0$. That is, $D^n = 0$. Note also that $D^{n-1}(x^{n-1}) = (n-1)! \neq 0$. Therefore $D^{n-1} \neq 0$. This tells us that the minimal polynomial of $D$ is $m(\lambda) = \lambda^n$.
Thus the Jordan form of $D$ consists of a single Jordan block of size $n$ corresponding to the eigenvalue $0$.