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!

Two endomorphisms are equal if and only if they act the same on all elements of a basis. Now
$$
DE(x^n)=D\left(\frac{1}{n+1}x^n\right)=x^n
$$
so indeed $DE=I$. However,
$$
ED(x^n)=E(nx^{n-1})=x^n
$$
if and only if $n>0$. Indeed, $ED(1)=E(0)=0\ne1$.
Next, prove by induction that $D^n\ne0$.
The base step is clear, because $D(x)=1$. Suppose $D^n\ne0$. Then take $f$ with $D^n(f)\ne0$. So $D^{n+1}(E(f))=D^nDE(f)=D^n(f)\ne0$. So $D^{n+1}\ne0$ as well.
