This is my attempt at proving the following: Suppose $A\overrightarrow{x}=\lambda\overrightarrow{x}$, then it follows that $A^n\overrightarrow{x}=\lambda^n\overrightarrow{x}$ where $n$ is a positive integer, $\lambda$ is some real number, and $A$ is a matrix.
Proof: We attempt a proof by induction, beginning to prove the case for $k=2$:
\begin{aligned} A^2\overrightarrow{x} &= A(A\overrightarrow{x})\\ &=A(\lambda\overrightarrow{x}) \\ &= \lambda(A\overrightarrow{x}) \hspace{3mm} \text{(as $\lambda$ is a scalar)} \\ &=\lambda^2\overrightarrow{x} \end{aligned} We now assume the result is true for some $n=k$ and use this to show it holds true for $n=k+1$. Thus, we have that \begin{aligned} A^k\overrightarrow{x} &=\lambda^k\overrightarrow{x} \\ A^{k+1}\overrightarrow{x} &=\lambda^k(A\overrightarrow{x}) \\ &=\lambda^{k+1}\overrightarrow{x} \end{aligned} By the induction hypothesis, we are done.
Is this proof ok?