Proofs involving endomorphisms on the space of polynomials Define endomorphisms $D$ and $E$ on the space of polynomials with rational co-efficients $ \mathbb{Q}[x] $ such that $ D(x^n)= nx^{n-1}, E(x^n) = \frac{1}{n+1}x^{n+1} $
We must show that $ DE = I $ but $ ED \neq I $
I attempted to write as a composition or as 2 matrices but got nowhere . Any ideas?
I also need to show that given $ f(x) \in \mathbb{Q}[x] $, show that $ D^n f(x) = 0 $, for some $ n \geq 1 $  
I really have no idea how to approach this, possibly because I haven't got part (i).
And finally given  $ n \geq 1 $, show $ D^n \neq 0 $. If someone could show me the method of answering this problem, I would appreciate it. Thanks!
If someone could show me the method of answering these questions that would be great. Thanks!

So the best way to see that $ED\neq I$ is to show that it is not injective, or that it has a nontrivial kernel.
Note: let $f(x) = 1$ be the constant polynomial. Then $D(f(x)) = 0$ (this should be part of the definition of $D$), so $ED(f(x)) = 0 \neq f(x)$. Thus, $ED\neq  I$.
In terms of your other questions, pursue this same thought : 
$D(c) = 0$ for $c\in \mathbb Q$. This will let you see that for each $f(x)$ there is some $n>0$ such that $D^n(f(x))=0$.
Yet for each $n>0$, what is $D^n(x^{n+1})$?
