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The exponential series is absolutely norm-convergent, since ${\sum }_{n\ge 1}\Vert x^n\Vert /n!\le e^{\Vert x\Vert }-1$. Completeness defines the Banach algebra exponential $e^x\in A$. If $xy=yx$, absolute convergence permits rearrangement of the product series, and the binomial identity givesTaking $y=-x$ yields $(e^x{)}^{-1}=e^{-x}$.

For the spectral hypothesis, put $K={\sigma }_A(u)$. Its complement has an open, hence path-connected, unbounded component containing $0$. Join $0$ to an exterior point by a simple polygonal arc in that component, and continue it to infinity, avoiding $K$. Removing this polygonal slit gives a simply connected open neighborhood $\Omega$ of $K$ on which $z$ has a branch of the complex logarithm $L(z)$. The holomorphic functional calculus therefore defines $x=L(u)$ and its composition rule givesThe relevant slit is chosen through the resolvent, not assumed to be a fixed negative-real-axis cut. This is the logarithm from a spectral slit in a Banach algebra.

Let $H$ be the subgroup of finite products of exponentials. It is a subgroup because $e^a{e}^{-a}=1$ and reversing a product gives its inverse. Each product is joined to $1$ by $t\mapsto e^{t{a}_1}\cdots e^{t{a}_m}$, so $H\subseteq G_0$. Conversely, if $\Vert v-1\Vert <1$, the convergent logarithm seriessatisfies $e^{\log v}=v$, by the scalar analytic identity and functional calculus. Thus $H$ contains a neighborhood of $1$. Its translates make $H$ open, and all other cosets are open too, making $H$ closed. Connectedness forces $G_0\subseteq H$. Finally $g{e}^a{g}^{-1}=e^{ga{g}^{-1}}$, so $H$ is normal. Thereforean open-and-closed normal subgroup. This is the identity component of Banach-algebra invertibles; finite products are essential and are not asserted to be single exponentials.