Nilpotent Jacobson radical I have to prove the following: If $R$ is a finite-dimensional algebra over a field $F$, then $J(R)$ is nilpotent.
I thought about this, but there are some gaps:  
Because $R$ is in particular a finite dimensional vector space, there is a composition series $0=R_0\subset R_1\subset\dots\subset R_k=R$ of VECTOR SPACES. Do I know that this is a composition series of $R$-modules? If yes, why? Or how can I find some?
If that is the case, I would proceed as follows: $J(R)R_n/R_{n-1}\subseteq R_n/R_{n-1}$. Now, I would like to use, that the quotient is simple, so if the above is a strict inclusion (how to show that?), then $J(R)R_n/R_{n-1}=0$. Now, by induction it is easy to see that $J(R)^k=\{0\}$. 

We can begin by saying $J=J(R)$ is finitely generated. The descending sequence of $J\supseteq J^2\supseteq \ldots$ must stabilize eventually, so at some point we have $J^n=J^{n+1}$. 
But the noncommutative formulation of Nakayama's Lemma says that if $M$ is a finitely generated right $R$ module $I$ is an ideal contained in $J(R)$ and $MI=M$, then $M=\{0\}$.
Applying that to this case, we have that $J^nJ=J^n$, so $J^n=\{0\}$.

I would like to use, that the quotient is simple, so if the above is a strict inclusion (how to show that?)

You have difficulties right away because it is unclear what $J(R)R_n/R_{n-1}$ means. It is not obvious that there should be a module structure on the vector space $R_n/R_{n-1}$.
But you do not have to use only vector spaces: a finite dimensional algebra is also right and left Artinian, so you could consider $R_n$'s to be right $R$ modules, and that $R_n/R_{n-1}$ is a simple right $R$ module.
At any rate, $R_{n-1}/R_{n-1}=J(R)(R_n/R_{n-1})\subsetneq R_n/R_{n-1}$ is a proper containment because $J(R)$ annihilates simple $R$ modules, including $R_n/R_{n-1}$.
