If $\frac{1}{n}A^n = \frac{1}{n+1}A^{n+1} = \frac{1}{n+2}A^{n+2}$ then any eigenvalue of $A$ is $0$, please check my proof I'd like to show that for a square matrix A with the property that there exists a positive integer n such that $\dfrac{1}{n}A^n = \dfrac{1}{n+1}A^{n+1} = \dfrac{1}{n+2}A^{n+2}$, any eigenvalue of A must be 0.
I proceed as follows: 
Let $\lambda$ be an eigenvalue of A, and $v$ be an eigenvector of A, such that $v \neq 0$.
Recall: $A^nv=\lambda^nv$, then since 
\begin{align} 
\dfrac{1}{n}A^n &=\dfrac{1}{n+1}A^{n+1} \\ 
\Rightarrow  \dfrac{n+1}{n}A^nv-A^{n+1}v &=0 \\
\Rightarrow \lambda^n\left(\dfrac{n+1}{n}-\lambda\right)v &= 0 \\
\end{align}
$\Rightarrow \lambda = 0$ or $\lambda = \dfrac{n+1}{n}$
Similarly: 
\begin{align} 
\dfrac{1}{n+1}A^{n+1} &=\dfrac{1}{n+2}A^{n+2} \\ 
\Rightarrow  \dfrac{n+2}{n+1}A^{n+1}v-A^{n+2}v &=0 \\
\Rightarrow \lambda^{n+1}\left(\dfrac{n+2}{n+1}-\lambda\right)v &= 0 \\
\end{align}
$\Rightarrow \lambda = 0$ or $\lambda = \dfrac{n+2}{n+1}$
My conclusion is since it's mathematically impossible that $\lambda = \dfrac{n+1}{n} = \dfrac{n+2}{n+1}$, the only eigenvalue of A must be 0. 

Is my demonstration incomplete or incorrect? 


Essentially correct. You should stick to mathematical english or to $\Rightarrow$ language, though and not mix them up as you do in the first block of the proof.
