I have seen this post here on math stack discussing the proof that the exponential of the adjoint matrix is the adjoint of the exponential of the original matrix, that is $\exp(A^{*}) = \exp(A)^{*}$. I would like to know if my alternative proof is correct.
My proof goes like this: The operation $T: A \mapsto A^{*}$, which takes whatever matrix and sends it to its adjoint is continuous in operator norm topology. This is true because, since $\|A^{*}\| = \|A\|$, $$\|A^{*} - B^{*}\| = \|(A-B)^{*}\| = \|A-B\|.$$ Thus, if we define: $$S_{N} = \sum_{n=0}^{N}\frac{1}{n!}A^{n}$$ we hahve $S_{N} \to \exp(A)$ and: $$TS_{N} = \sum_{n=0}^{N}\frac{1}{n!}T(A^{n}) = \sum_{n=0}^{N}\frac{1}{n!}(A^{*})^{n}.$$ By continuity: $$\sum_{n=0}^{N}\frac{1}{n!}A^{n} \to \exp(A) \Rightarrow TS_{N} = \sum_{n=0}^{N}\frac{1}{n!}(A^{*})^{n} \to TS_{\infty} = \exp(A)^{*}$$ and the result follows.
Is my proof correct?