If $A,B \in \mathcal{M}_n$ and $AB=BA=\mathbb{O}_n$ prove that $(A+B)^ν = Α^ν + Β^ν$. I have one exercise in Linear Algebra and I would like to know if my solution is correct.
If $A,B \in \mathcal{M}_n$ and $AB=BA=\mathbb{O}_n$ prove that $(A+B)^ν = Α^ν + Β^ν$.

My first thought is that we cant do the binomial theorem on the $(A+B)^ν$.So we must find another path to solve the problem.If we take $AB=\mathbb{O}_n \Rightarrow A^{-1}AB=\mathbb{O}_n \Rightarrow B=\mathbb{O}_n$ and $BA=\mathbb{O}_n \Rightarrow B^{-1}BA=\mathbb{O}_n \Rightarrow A=\mathbb{O}_n$ then $(A+B)^ν = Α^ν + Β^ν \Leftrightarrow (\mathbb{O}_n+\mathbb{O}_n)^ν=\mathbb{O}_n^ν + \mathbb{O}_n^ν \Leftrightarrow \mathbb{O}_n=\mathbb{O}_n$ so its true.
Is my thought correct?

I'm assuming $\mathbb O_n$ means the $n \times n$ matrix $0$ (usually just written as $0$, sometimes $\bf 0$).
Actually the binomial theorem does work when $A$ and $B$ commute, which is the case here.  The proof is the same as for the case of numbers. 
Your attempted proof, however, is wrong.  You can have $A B = B A = 0$ without $A = B = 0$, e.g. try $ \pmatrix{1 & 0\cr 0 & 0\cr}$ and $\pmatrix{0 & 0\cr 0 & 1\cr}$.
