Question: > 
> Let $A$ and $B$ be two nilpotent matrices. Prove that $A+B$ is nilpotent.
> 
> 
> 

I'm new to the study of matrices, I already proved that $AB$ is nilpotent too (that was the previous question) but I have no clue about solving this one.
Plus, I was wondering if I have to prove that it exists an $n$ such that $(A+B)^n=0$ or such that $A^n+B^n=0$.

Any help?

EDIT:

I just know that:
$(A+B)^k=(A+B)(A+B)...(A+B)$ $k$ times, so I need to prove that $(A+B)=0$ 

Is this enough to prove what I need, right?

Answer: 
Assume that there are $m,n\in\mathbb{N}$ such that $A^m=0,B^n=0.$

Then 
$$(A+B)^{m+n}=\sum\_{k=0}^{m+n} C\_{m+n}^k A^{m+n-k}B^{k}$$
provided AB=BA.

When $k<n$, we know $m+n-k>m$, which means $A^{m+n-k}=0$. And when $k\geq n$, we know $B^k=0.$
That is $(A+B)^{m+n}=0,$ which means A+B is nilpotent.

But I have no idea how to answer it if $AB\neq BA$.
I think you're missing a hypothesis: the group of the matrices has to be commutative! Otherwise the thesis doesn't hold:

let 
$
A=
\begin{bmatrix} 
0 & 1 \\
0 & 0 
\end{bmatrix}$ and 
$
B=
\begin{bmatrix} 
0 & 0 \\
1 & 0 
\end{bmatrix}$

We have that $A^2=0$ and $B^2=0$, but $(A+B)^2=
\begin{bmatrix} 
1 & 0 \\
0 & 1 
\end{bmatrix}$
 which is definitely not the zero matrix!