[proofplan]
We prove the local inverse property directly from absolutely convergent [power series](/page/Power%20Series) in the finite-dimensional normed algebra $M(n,\mathbb C)$. First, the submultiplicative norm gives absolute convergence of the exponential and logarithm series and allows Cauchy-product manipulations for one matrix at a time. The scalar formal identities $\log(\exp z)=z$ near $0$ and $\exp(\log(1+z))=1+z$ for $|z|<1$ then transfer to matrices because all powers of a fixed matrix commute. Finally, we choose explicit norm balls around $0$ and $I_n$ so that the logarithm is defined and the matrices in the logarithm neighbourhood are invertible by the Neumann series.
[/proofplan]
[step:Establish absolute convergence for the exponential and logarithm series]
Let $X \in M(n,\mathbb C)$. Since $\|\cdot\|_*$ is submultiplicative, induction gives $\|X^k\|_* \leq \|X\|_*^k$ for every integer $k \geq 0$. Hence
\begin{align*}
\sum_{k=0}^{\infty}\left\|\frac{X^k}{k!}\right\|_* \leq \sum_{k=0}^{\infty}\frac{\|X\|_*^k}{k!}=\exp(\|X\|_*).
\end{align*}
Thus the matrix exponential is well-defined for every $X \in M(n,\mathbb C)$.
Let $A \in M(n,\mathbb C)$ satisfy $\|A\|_*<1$. Again using submultiplicativity,
\begin{align*}
\sum_{k=1}^{\infty}\left\|(-1)^{k+1}\frac{A^k}{k}\right\|_* \leq \sum_{k=1}^{\infty}\frac{\|A\|_*^k}{k}.
\end{align*}
The scalar series on the right converges for $\|A\|_*<1$. Therefore the logarithm series
\begin{align*}
\sum_{k=1}^{\infty}(-1)^{k+1}\frac{A^k}{k}
\end{align*}
converges absolutely whenever $\|A\|_*<1$, and $\log Y$ is well-defined for every $Y$ with $\|Y-I_n\|_*<1$.
[guided]
The norm condition is what permits us to treat the matrix series like scalar power series with controlled coefficients. For the exponential, submultiplicativity gives the estimate $\|X^k\|_* \leq \|X\|_*^k$ for every $k \geq 0$. Therefore the matrix series is dominated term-by-term by the scalar exponential series:
\begin{align*}
\sum_{k=0}^{\infty}\left\|\frac{X^k}{k!}\right\|_* \leq \sum_{k=0}^{\infty}\frac{\|X\|_*^k}{k!}.
\end{align*}
The scalar series converges to $\exp(\|X\|_*)$, so the matrix exponential series converges absolutely in the finite-dimensional [normed vector space](/page/Normed%20Vector%20Space) $M(n,\mathbb C)$.
For the logarithm, write $A=Y-I_n$. The logarithm series is a power series in the single matrix $A$. If $\|A\|_*<1$, then
\begin{align*}
\sum_{k=1}^{\infty}\left\|(-1)^{k+1}\frac{A^k}{k}\right\|_* \leq \sum_{k=1}^{\infty}\frac{\|A\|_*^k}{k}.
\end{align*}
The scalar series $\sum_{k=1}^{\infty}\|A\|_*^k/k$ converges because $\|A\|_*<1$. Hence the logarithm series is absolutely convergent. This proves that both series used in the theorem are legitimate matrix-valued expressions on the stated domains.
[/guided]
[/step]
[step:Choose neighbourhoods where the logarithm is defined and matrices are invertible]
Define
\begin{align*}
U:=\{X \in M(n,\mathbb C):\|X\|_*<\log 2\}
\end{align*}
and
\begin{align*}
V:=\{Y \in M(n,\mathbb C):\|Y-I_n\|_*<1\}.
\end{align*}
The set $U$ is an open neighbourhood of $0$ in $M(n,\mathbb C)$.
We next show that $V \subset GL(n,\mathbb C)$. Let $Y \in V$ and define $A:=Y-I_n$. Then $\|A\|_*<1$. The Neumann series
\begin{align*}
S:=\sum_{j=0}^{\infty}(-A)^j
\end{align*}
converges absolutely, since
\begin{align*}
\sum_{j=0}^{\infty}\|(-A)^j\|_* \leq \sum_{j=0}^{\infty}\|A\|_*^j<\infty.
\end{align*}
For each integer $m \geq 0$, define the partial sum
\begin{align*}
S_m:=\sum_{j=0}^{m}(-A)^j.
\end{align*}
Then
\begin{align*}
(I_n+A)S_m=I_n+(-A)^{m+1}
\end{align*}
and
\begin{align*}
S_m(I_n+A)=I_n+(-A)^{m+1}.
\end{align*}
Since $\|(-A)^{m+1}\|_* \leq \|A\|_*^{m+1}\to 0$, the right-hand side $I_n+(-A)^{m+1}$ converges to $I_n$ in the norm $\|\cdot\|_*$. Also $S_m \to S$ by definition of the absolutely convergent series. Multiplication by a fixed matrix is continuous with respect to $\|\cdot\|_*$ because submultiplicativity gives
\begin{align*}
\|(I_n+A)S_m-(I_n+A)S\|_* \leq \|I_n+A\|_*\|S_m-S\|_*
\end{align*}
and similarly
\begin{align*}
\|S_m(I_n+A)-S(I_n+A)\|_* \leq \|S_m-S\|_*\|I_n+A\|_*.
\end{align*}
Passing to the limit in both finite identities gives $(I_n+A)S=S(I_n+A)=I_n$. Thus $Y=I_n+A$ is invertible. Hence $V \subset GL(n,\mathbb C)$, and $V$ is an open neighbourhood of $I_n$ in $GL(n,\mathbb C)$ with $V \subset \{Y:\|Y-I_n\|_*<1\}$.
[/step]
[step:Prove the exponential and logarithm identities for one matrix by scalar power-series composition]
We use the following scalar power-series identities:
\begin{align*}
\log(\exp z)=z
\end{align*}
for all complex $z$ sufficiently close to $0$, and
\begin{align*}
\exp(\log(1+z))=1+z
\end{align*}
for all complex $z$ with $|z|<1$. These identities hold as identities of absolutely convergent scalar power series on the indicated discs.
Let $X \in U$. Define
\begin{align*}
B:=\exp X-I_n.
\end{align*}
Using the estimate from the first step,
\begin{align*}
\|B\|_* \leq \sum_{k=1}^{\infty}\frac{\|X\|_*^k}{k!}=\exp(\|X\|_*)-1<1,
\end{align*}
because $\|X\|_*<\log 2$. Hence $\log(\exp X)$ is defined.
The series defining $B$ is a power series in the single matrix $X$, and all powers of $X$ commute with each other. Let $r:=\|X\|_*$. Since $r<\log 2$, the scalar quantity $e^r-1$ is less than $1$, and
\begin{align*}
\sum_{m=1}^{\infty}\frac{1}{m}\left(\sum_{k=1}^{\infty}\frac{r^k}{k!}\right)^m=\sum_{m=1}^{\infty}\frac{(e^r-1)^m}{m}<\infty.
\end{align*}
For each fixed integer $m\geq 1$, the $m$-fold Cauchy product
\begin{align*}
\left(\sum_{k=1}^{\infty}\frac{X^k}{k!}\right)^m
\end{align*}
is absolutely dominated by
\begin{align*}
\left(\sum_{k=1}^{\infty}\frac{r^k}{k!}\right)^m.
\end{align*}
The preceding summability therefore gives absolute convergence of the full double composition and permits regrouping terms by total degree. After regrouping, the coefficient of $X^q$ in this matrix power series is exactly the coefficient of $z^q$ in the scalar power series $\log(\exp z)$, since every product of powers of $X$ reduces to a single power $X^q$. The scalar identity $\log(\exp z)=z$ holds as an identity of convergent power series near $0$, so the coefficient is $1$ for $q=1$ and $0$ for every $q \neq 1$. Hence
\begin{align*}
\log(\exp X)=X.
\end{align*}
Now let $Y \in V$ and define $A:=Y-I_n$. Then $\|A\|_*<1$, so $\log Y$ is defined by the absolutely convergent series
\begin{align*}
L:=\log Y=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{A^k}{k}.
\end{align*}
The matrix $L$ is a power series in the single matrix $A$, so powers of $L$ expand as absolutely convergent Cauchy products of powers of $A$. The scalar majorant is the absolutely convergent composition $\exp(\log(1+z))$ for $|z|<1$. Therefore the coefficient of $A^q$ in
\begin{align*}
\exp L=\sum_{m=0}^{\infty}\frac{L^m}{m!}
\end{align*}
is the coefficient of $z^q$ in the scalar power series $\exp(\log(1+z))$. Since $\exp(\log(1+z))=1+z$ for $|z|<1$, the only nonzero coefficients are the constant coefficient $1$ and the coefficient $1$ of $z$. Thus
\begin{align*}
\exp(\log Y)=I_n+A=Y.
\end{align*}
[guided]
The point of this step is that no noncommutative difficulty occurs: every expression is a power series in one matrix. For $\log(\exp X)$, the relevant matrix is $X$. Since $X^aX^b=X^{a+b}$, multiplying power series in $X$ has exactly the same coefficient bookkeeping as multiplying scalar power series in one variable.
First let $X \in U$, so $\|X\|_*<\log 2$. Define $B:=\exp X-I_n$. The exponential estimate gives
\begin{align*}
\|B\|_* \leq \sum_{k=1}^{\infty}\frac{\|X\|_*^k}{k!}=\exp(\|X\|_*)-1.
\end{align*}
Because $\|X\|_*<\log 2$, the last quantity is strictly less than $1$. Hence the logarithm series for $\log(I_n+B)=\log(\exp X)$ is defined.
Now consider the composition
\begin{align*}
\log(\exp X)=\sum_{m=1}^{\infty}(-1)^{m+1}\frac{B^m}{m}.
\end{align*}
Substituting $B=\sum_{k=1}^{\infty}X^k/k!$ gives a power series in $X$. We justify the Cauchy-product multiplication by a scalar majorant. Put $r:=\|X\|_*$. Since $r<\log 2$, we have $\exp r-1<1$, and therefore
\begin{align*}
\sum_{m=1}^{\infty}\frac{1}{m}\left(\sum_{k=1}^{\infty}\frac{r^k}{k!}\right)^m=\sum_{m=1}^{\infty}\frac{(\exp r-1)^m}{m}
\end{align*}
converges. For each fixed $m$, the norm of the $m$-fold Cauchy product in $X$ is bounded by the corresponding $m$-fold scalar Cauchy product in $r$. Summing over $m$ gives absolute convergence of the whole composition. Thus regrouping the terms by total degree is valid.
After this justification, the coefficients are the same as in the scalar identity. The scalar power series identity near $0$ is
\begin{align*}
\log(\exp z)=z.
\end{align*}
Therefore the coefficient of $X$ is $1$, and every coefficient of $X^q$ for $q \neq 1$ is $0$. This proves
\begin{align*}
\log(\exp X)=X.
\end{align*}
For the second identity, let $Y \in V$ and set $A:=Y-I_n$. Then $\|A\|_*<1$. Define
\begin{align*}
L:=\log Y=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{A^k}{k}.
\end{align*}
Again $L$ is a power series in the single matrix $A$, so all products appearing in $\exp L$ are products of powers of $A$ and commute. The absolute convergence is controlled by the scalar series for $\exp(\log(1+z))$ at $|z|=\|A\|_*<1$. Hence the matrix coefficients agree with the scalar coefficients. Since
\begin{align*}
\exp(\log(1+z))=1+z
\end{align*}
for $|z|<1$, the matrix identity becomes
\begin{align*}
\exp(\log Y)=I_n+A=Y.
\end{align*}
[/guided]
[/step]
[step:Combine the neighbourhood choices with the two identities]
The set $U=\{X:\|X\|_*<\log 2\}$ is an open neighbourhood of $0$ in $M(n,\mathbb C)$, and the set $V=\{Y:\|Y-I_n\|_*<1\}$ is an open neighbourhood of $I_n$ contained in $GL(n,\mathbb C)$. The previous step proves that $\log(\exp X)=X$ for every $X \in U$ and $\exp(\log Y)=Y$ for every $Y \in V$. These are precisely the required local inverse identities for the matrix exponential and the power-series logarithm.
[/step]
