Exercise 22

Show that if $\lambda$ is an eigenvalue of $A$ then $\lambda^{k}$ is an eigenvalue of $A^{k}$.

What is wrong with this proof generalizing that? “If $\lambda$ is an eigenvalue of $A$ and $\mu$ is an eigenvalue for $B$, then $\lambda\mu$ is an eigenvalue for $AB$, for, if $A\vec{x}=\lambda\vec{x}$ and $B\vec{x}=\mu\vec{x}$ then $AB\vec{x}=A\mu\vec{x}=\mu A\vec{x}=\mu\lambda\vec{x}$”?