I'm looking over the following proof, and I'm unsure of one particular part.
Theorem: Suppose that $a \in A$ is a Hermitian element of a $C^{*}-$Algebra $A$. Then $\sigma(a) \subset \mathbb{R}$.
Proof: WLOG We may assume that $A$ is unital, otherwise we can prove the result for the unitization $\hat{A}$ and use the fact that $\sigma_{\hat{A}}(a) = \sigma_A(a)$. Since $e^{ia}$ is unitary, $\sigma(e^{ia}) \subset T$ for $T$ the unit circle in $\mathbb{C}$. If $\lambda \in \sigma(a)$, and $b = \sum_{n=1}^{\infty} \frac{i^n(a-\lambda I)^{n-1}}{n!}$, then: $$e^{ia}-e^{i\lambda}I = (e^{i(a-\lambda I)}-I)e^{i \lambda} = (a-\lambda I)be^{i \lambda}.$$ Since $b$ commutes with $a$ and $(a-\lambda I) \notin \text{Inv}(A)$, it follows that $e^{ia}-e^{i\lambda}I \notin \text{Inv}(A)$, so $e^{i \lambda} \in \sigma(e^{ia}) \subset T$, so $\lambda \in \mathbb{R}$.
The proof is simple enough, but I don't understand why we need $b$ to commute with $a$. Like, I understand why $b$ commutes with $a$, since $a$ commutes with $(a-\lambda I)^n$ for all $n \in \mathbb{N}$. But this fact seems meaningless for the proof to work.
Isn't it always the case that If I have a product of elements $\prod_{k=1}^n a_k \in A$ in an algebra $A$ that $\prod_{k=1}^n a_k \in \text{Inv}(A)$ if and only if $a_k \in \text{Inv}(A)$ for all $k=1,2,\dots,n$? Do the $a_k$'s have to commute with eachother too?
So wouldn't we simply need to state: "Since $(a-\lambda I) \notin \text{Inv}(A)$, the product $(a-\lambda I)be^{i \lambda} = e^{ia} - e^{-\lambda}I \notin \text{Inv}(A)$."?
No, it's not true in general. The typical example is the unilateral shift $S$, where you have $S^*S=1$ but $S$ is not invertible since it is not surjective.