Let $\mathbf{E}$ be a real banach space and $\mathbf{L} = L(\mathbf{E})$ the resultant banach space of bounded linear operators $T:\mathbf{E} \to \mathbf{E}$ equipped with the operator norm $\| T \| = \sup_{\| x \| \leqslant 1 } \|Tx\|$.
From https://en.wikipedia.org/wiki/Neumann_series I can see why the subset $\Omega \subset \mathbf{L}$ consisting of all invertible operators is an open set of $\mathbf{L}$. Moreover, from the accepted answer to differential inverse matrix I can see why the map $\Phi: \Omega \mapsto \mathbf{L} \, \,; \,\, T \mapsto T^{-1}$ is frechet differentiable, and specifically $D\Phi(T)S = -T^{-1} S T^{-1}$.
How do I use this knowledge to prove the stronger result that $\Phi$ is smooth?
Regarding finite dimensional spaces $\mathbf{E}$, this question has been asked before on stackexchange, but the answers almost always employ Cramer's rule for matrices and partial derivatives as ratios of smooth functions. I am not assuming $\mathbf{E}$ is finite dimensional, so Cramer's rule doesn't apply. I'm trying to find an answer that doesn't deal with partial derivatives at all.
As seen from your link to the Neumann series, the inversion map $\Phi$ can be given by a power series at $T$ as: $$ \Phi(T+S) = \sum_{n \ge 0} (-T^{-1} S)^n T^{-1}. $$ This power series has radius of convergence $\rho = \frac{1}{\|T^{-1}\|}>0$, so it absolutely converges inside that, and hence $\Phi$ is analytic.