Nilpotent commutative matrices $A, B$ $\Longrightarrow$ $A+B$ nilpotent. How to prove that if $A, B$ are matrix of $n\times n$ nilpotents so that $AB=BA$ then $A+B$ is nilpotent.
 A: If $A^m = 0$ and $B^n = 0$ then what is $(A+B)^{m+n}$? Use the binomial theorem (uses commutativity!)
Edit: I'll give you one more step.
$$
(A+B)^{m+n} = \sum_{i=0}^{m+n} {m+n \choose i}A^{i}B^{m+n-i}
$$
can you show that either $A^i = 0$ or $B^{m+n-i}=0$?
A: Another approach to the problem can be based on the well known fact that commuting matrices are simultaneously triangularizable.
If so then exists such matrix $P$ that $A=PT_AP^{-1}$ and $B=PT_BP^{-1}$ where $T_A,T_B$ are, assume, upper triangular matrices.    
Additionally they have on diagonal only $0$ values as the eigenvalues of nilpotent matrices are $0$. 
Hence it's obvious that $T_A+T_B$ has also only eigenvalues equal to $0$ and consequently also $A+B=  PT_AP^{-1} +PT_BP^{-1}=P(T_A+T_B)P^{-1}$
This kind of argumentation allows additionally to extend the claim for other types of expressions with commuting nilpotent $A,B$ so we have for example that    
any sums of powers $A^k, B^i$ $(k,i>0)$ (and their products) are also nilpotent.
