Spectrum of the operator Let $T$ be an operator on Hilbert space. 
Define $\sigma(T)=\lbrace \lambda\in \mathbb{C} | \lambda I - T~\textrm{is not invertible}\rbrace$. 
How can I prove that $\sigma(T^n)=\lbrace \lambda^n|\lambda\in \sigma(T)\rbrace$?
 A: Note: Given a polynomial $P(x)$, we can talk about the operator $P(A)$ which is the polynomial applied to $A$.  Since $A$ commutes with itself, and the identity commutes with everything, we can factor $P(A)=(A-c_1 I)(A-c_2 I)\cdots (A-c_m I)$ by the fundamental theorem of algebra.  
Hint: The polynomial $x-\lambda$ divides $x^n-\lambda^n$.  This shows $\{\lambda^n:\ \lambda\in \sigma(A)\} \subset \sigma(A^n)$.  To show that each element $r\in\sigma(A^n)$ is of the form $\lambda^n$, $\lambda\in \sigma(A)$, consider $x^n-r$, and factor this over $\mathbb{C}$ as a polynomial.  Since the polynomial evaluated at $A$ is non-invertible, at least one factor must non-invertible.  Hence $r^{1/n}\zeta_n\in \sigma(A)$ where $\zeta_n$ is some root of unity, and the proof is finished.
Hope that helps,
Edit:  More generally, suppose we are given an operator $A$, and that $Q(x)=\frac{p(x)}{q(x)}$ is a rational function whose poles do not lie in $\sigma(A)$.  Then $Q(A)$ makes sense as an operator, and $$\sigma(Q(A))=\{Q(\lambda):\ \lambda\in\sigma(A)\}.$$  The proof of this is very similar to the proof of what you want.  
Edit 2:  Since we are talking about the spectrum, I removed the earlier edit.  I'll still mention, the exact same proof works with the directions changed to show that the resolvent is preserved under polynomials.
