$\sigma(T^n)=[\sigma(T)]^n$ Let $T$ be a bounded linear operator on a Hilbert space $\mathcal{H}$. The spectrum $\sigma(T)$ is defined to be the set of all $\lambda\in\mathbb{C}$ that makes $\lambda I-T$ not invertible.
I want to see that $\sigma(T^n)=\left[\sigma(T)\right]^n$, where $n\geq 1$ and $A^n=\{a^n:a\in A\}$ for any $A\subset\mathbb{C}$.
(This is Exercise V.4 in Retherford's Hilbert space: compact operators and the trace theorem.
One way is straightforward: if $\lambda\in\sigma(T)$ so that $\lambda I-T$ is not invertible, then $\lambda^nI-T^n=(\lambda I-T)(\lambda^{n-1}I+\lambda^{n-2}T+\cdots+T^{n-1})$ shows that $\lambda^nI-T^n$ is not invertible, i.e., $\lambda^n \in\sigma(T^n)$.
The way I am stuck is the converse, mostly because a spectral element need not be an eigenvalue in an infinite-dimensional case.
 A: I guess your first attempt is wrong (you want to show $\sigma(T^n) \subseteq [\sigma(T)]^n$ while you show something trivial).
Consider the factorization (which exists by fundamental theorem of algebra):
\begin{align*}
\lambda - z^n = \prod_{j = 1}^n (z - \lambda_j) \qquad (z \in \mathbb C).
\end{align*}
"$\subseteq$": Let $\lambda \in \sigma(T^n)$. Then $\lambda - T^n$ is not invertible and due to the factorization there is a $j \in [1:n]$ such that $\lambda_j - T$ is not invertible and thus $\lambda_j \in \sigma(T)$. It follows $\lambda = {\lambda_j}^n \in [\sigma(T)]^n$.
"$\supseteq$": Let $\lambda \in [\sigma(T)]^n$. Then there is a $\mu \in \sigma(T)$ such that $\lambda = \mu^n$ or equivalently $\lambda - \mu^n = 0$. Hence there is a $k \in [1:n]$ such that $\mu = \lambda_k$. Suppose that $\lambda \not \in \sigma(T^n)$. Then $\lambda - T^n$ is invertible and thus 
\begin{align*}
I = (\lambda - T^n)^{-1} \prod_{j = 1}^n (T - \lambda_j) = (\lambda - T^n)^{-1} \prod_{j = 1 \atop j \neq k}^n (T - \lambda_j) (T - \lambda_k) 
\end{align*}
using the fact that the operators $T - \lambda_j$ commute pairwise. It follows that $\lambda_k - T$ has a left inverse and analoguely one can show that $\lambda_k - T$ has a right inverse and therefore $\lambda_k \in \rho(T)$. That contradicts the fact that $\lambda_k = \mu \in \sigma(T)$. Hence it follows $\lambda \in \sigma(T^n)$. 
If you want to go further with these kinds of exercises you can show with an analogue proof (that additionally deals with some trivial cases) that it even holds that $\sigma(p(T)) = p(\sigma(T))$ for all polynomials $p \in \mathbb C[z]$. 
