Note: Please do not give a solution; I would prefer guidance to help me complete the question myself. Thank you.
Show that if $S,T\in\mathcal{L}(X)$, where $X$ is a Banach space over $\mathbb{C}$. Then $\sigma(ST)\setminus\{0\}=\sigma(TS)\setminus\{0\}$.
I have tried to look at the Neumann Series for $(I-ST)^{-1}$ as well as using the definition of the resolvent set, $\rho(ST)$, to try and show something about the resolvent, $R(\lambda,ST)$.
Please, as I have noted above. Do not include any solutions. Just helpful guidance. I am studying spectral theory and there is no point being given an answer otherwise reading texts without exercise questions would be sufficient to learn. I appreciate all help. Finally, for some context, this is part of a functional analysis/PDE's course, so hopefully that will help people to understand what my frame of mind is coming to this question.
My Proof
First we show that $ST$ and $TS$ are continuous and hence a spectrum and resolvent set are well defined. Using $S,T\in\mathcal{L}(X)$, \begin{align} \|ST(x)\|\leq\|S\|\|Tx\|\leq\alpha\|S\|\|x\|\leq M, \end{align} for all $x\in X$ so $ST\in\mathcal{L}(X)$. Reversing the roles of $S$ and $T$ gives $TS\in\mathcal{L}(X)$.
Suppose $\lambda\in\rho(ST)$, then $\lambda I-ST$ is invertible. Let $U=(\lambda I-ST)^{-1}$. Now, \begin{align} \frac{1}{\lambda}(I+TUS), \end{align} is well defined for $\lambda\neq 0$. Then, \begin{align} \frac{1}{\lambda}(I+TUS) &= \frac{1}{\lambda}(I+T(\lambda I-ST)^{-1}S)\\ &= \frac{1}{\lambda}(I+(S^{-1}(\lambda I-ST)T^{-1})^{-1})\\ &= \frac{1}{\lambda}(I+(\lambda S^{-1}T^{-1}-I)^{-1}) = \frac{1}{\lambda}(I+(\lambda(TS)^{-1}-I)^{-1})\\ &= \frac{1}{\lambda}(I+((TS)^{-1}(\lambda I-TS))^{-1})\\ &= \frac{1}{\lambda}(I+(\lambda I-TS)^{-1}TS)\\ &= \frac{1}{\lambda}(\lambda I-TS)^{-1}((\lambda I-TS)+TS)\\ &= (\lambda I-TS)^{-1}. \end{align} Therefore, $(\lambda I-TS)^{-1}$ exists and hence $\lambda I-TS$ is invertible. That is, given $\lambda\in\rho(ST)$, such that $\lambda\neq 0$, then $\lambda\in\rho(TS)$. We can show the same for any $\lambda\in\rho(TS)$, such that $\lambda\neq 0$, then $\lambda\in\rho(ST)$ using the same procedure as above. Hence $\rho(ST)\setminus\{0\}=\rho(TS)\setminus\{0\}$ and therefore $\sigma(ST)\setminus\{0\}=\sigma(TS)\setminus\{0\}$ by definition.
If $I-ST$ is invertible with inverse $U$ then $I+TUS$ is an inverse for $I-TS$. This can be verified by simple algebra. To prove that $\sigma (ST)\setminus \{0\}=\sigma (ST)\setminus \{0\}$ what you have to prove is that if $\lambda \neq 0$ then $\lambda \notin \sigma (ST)$ iff $\lambda \notin \sigma (TS)$. Can you complete the proof now? (You may have to replace $T$ or $S$ in the first part by a scalar multiple).