If $\lambda$ is an eigenvalue of $A$, determine eigenvalues of $A^2$ and $A^3$ Also what is an eigenvalue of $A^n$?
I know you can square the eigenvalues, etc... but how do I prove this for a general case? 
 A: If $p(x)$ is a rational function defined on the spectrum of $A$ then the spectrum $$\sigma(p(A))=p(\sigma(A))$$
The spectrum being the set with all the eigenvalues or in a more precise way $\sigma(A)$={ $\lambda I-A$ is not invertible}. 
Such definition is because this theorem comes from operator theory where in principle $A$ doesn't have to be a matrix, but in you case you can take the ordinary definition of spectrum being all the eigen values of $A$. So in your special case: $\sigma(A^m)=(\sigma(A))^m$ where the last is an abuse of notation to indicate $\sigma(A)$= {$\lambda^n$ : $\lambda$ eigenvalue of $A$ }
If you'd like some demonstration, it does need a little bit of effort but essentially you can work it out starting from the "Jordan Canonical Form" noticing that the matrix $J$ which represent the canonical form of $A$ is the result of $J=D+N$ in other word is the result of a $D$ diagonal matrix and a $N$ nihilpotent matrix. Then you apply the function and you will work out more or less the theorem.
A: If $E$ is an eigenspace of $A$ correspoinding to an eigenvalue $\lambda$ then
$A$ acts as a multiplication by $\lambda$ operator on $E$, that is,
$A x = \lambda x$ if $x \in E$.
Note that $AE \subset E$, so you can apply $A$ again, where it still acts as a multiplication by $\lambda$ operator on $E$. Hence $A(A x) = A(\lambda x) = \lambda Ax = \lambda ^2 x$.
