An overkill way to recover the Taylor expansion of $e^x$

Let us consider the linear operator of Banach spaces $$F \colon C[0, \, 1/2] \to C[0, \, 1/2], \quad F(f(x)):= \int_0^x f(t)dt.$$ Then the identity $\int_0^x e^tdt=e^x-1$ can be rewritten as $$(I-F)(e^x)=1. \quad (\sharp)$$ The operator $F$ is bounded and we have $$||F|| = \sup_{f \in C[0, \, 1/2]} \frac{||F(f)||}{||f||} \leq \frac{1/2 \, ||f||}{||f||} = \frac{1}{2} <1.$$ This implies that $I-F$ is invertible as a bounded linear operator and  moreover $$(I-F)^{-1} = \sum_{k=0}^{+ \infty} F^k.$$ Substituting in $(\sharp)$, we get $$e^x=(I-F)^{-1}(1)=\left(\sum_{k=0}^{+ \infty} F^k \right) (1) = \sum_{k=0}^{+ \infty} F^k(1) = \sum_{k=0} ^{+ \infty}\frac{x^k}{k!},$$ which is the well-known Taylor series expansion of  $e^x$.