Basic logic problem with verbal question, confirmation whether right or wrong Problem 2a here on page 882, translated

Prove the statement
If $\lambda\in \sigma(A)$, so $\lambda^p \in\sigma(A^p) \forall
 p\in\mathbb N.$
(where $\lambda$ is an eigen-value and $\sigma(A)$ is a set of the eigen-values)

Suppose arbitrary $A$ so that $\lambda\in\sigma(A)$. By the definition of eigen-value, we have a non-zero $\bar{x}$ so that $A\bar{x}=\lambda \bar{x}$ where $\bar{x}\in\mathbb R^n$,  $\lambda\in\mathbb R$ and $A$ is some matrix such that $A\in R^{n\times m}$.

Diversion
  I earlier abserved here, $A\bar{x_1}=\lambda_1\bar{x_1}$ so $A^2\bar{x_1}=\lambda_1^2\bar{x_1}$
  so $A^k\bar{x_k}=\lambda_k^k\bar{x_k}$ so $\lambda_k\in \sigma(A^k)
\forall k\in \mathbb N$.

Now I think I must somehow use that, thinking (sorry about bad quality but it is only meant for a very fast draft) -- basically the junk means use indunction after quantifying the terms and prove with direct proof.


Goal We need to prove that $\forall p\in\mathbb N : \exists \bar{x}\in\mathbb R \not =\bar{0} : A^p \bar{x}=\lambda^p\bar{x}$.

 A: I don't understand why you would consider the identity matrix separately. 
There is absolutely no reason to do so.
(In addition, it is very bad form to begin a proof for a statement that has an implied universal quantifier (here, "for every matrix $A$") by saying "Let's choose...".  One cannot prove a universal statement by choosing a particular instance (unless the universe has one and only one element, in which case you aren't really choosing anything). If you are really going to deal with the issue by cases, then it is much better phrasing to begin by saying "Let $A$... Then either $A=[1]$ or $A\neq [1]$; if $A=[1]$ then ... If $A\neq[1]$, then ..." or some such.)
Remember the definition of "eigenvalue". Given a square matrix $B$, a scalar $\lambda$ is an eigenvalue of $B$ if and only if there exists a nonzero vector $\mathbf{x}$ such that $B\mathbf{x}=\lambda\mathbf{x}$.
If $\lambda\in \sigma(A)$, then that means that there exists a nonzero vector $\mathbf{v}$ such that $A\mathbf{v}=\lambda\mathbf{v}$.
In order to show that $\lambda^p\in\sigma(A^p)$, you need to show that there exists a nonzero vector $\mathbf{x}$ such that $A^p\mathbf{x}=\lambda^p\mathbf{x}$. The observation you make tells you which vector $\mathbf{x}$ you might want to select, and how to show that you do, in fact, get the desired equality.
If such a vector exists, then you are done: you've proven that if $\lambda\in\sigma(A)$, then necessarily $\lambda^p\in\sigma(A^p)$. 
Your solution to the general case is poorly presented: you never specify that $\mathbf{x}_1\neq\mathbf{0}$, and a $\mathbf{x}_k$ appears ex nihilo in the final clause. What is $\mathbf{x}_k$? 
