Proving the nth derivative of a polynomial of degree n-1 is zero using linear Algebra. I want to prove by using linear algebra only that the nth derivative of a polynomial of degree n-1 is zero. My idea is using proving first that every square matrix $A$ such the only not zero entries are those that $j=i+1$ then $A^{n}=0$.
Then by noticing that for the derivative operator $D:P_{n}(\mathbb{R}) \to P_{n}(\mathbb{R})$ and the canonical basis $\beta= \lbrace 1, x,x^{2},...,x^{n} \rbrace$ of the vector space $P_{n}(\mathbb{R})$ we have that $[D]_{\beta}$ is such the entry $j=i+1$ then $([D]_{\beta})^{n}=0$ so the nth derivaritve of a polynomial can be seen in a matrix representation as
$$ [D^{n}p(x)]_{\beta}=[D^{n}]_{\beta}[p(x)]_{\beta}=([D]_{\beta})^{n}[p(x)]_{\beta}=0 [p(x)]_{\beta}=0 $$
Proving that the  the nth derivaritve of a polynomial of degree $n-1$ is zero but this also proves thath the nth derivative of a polynomial of degrees $n$ is zero which is not true. So.. what I am doing wrong and end up the proof of this only using linear algebra? Thanks

Note that $[D]_{\beta}$ is an $(n+1)\times (n+1)$ matrix. What we actually have is that $([D]_{\beta})^{n+1} = 0$, not $([D]_{\beta})^n=0$. 
