[proofplan]
We separate the zeroth term because submultiplicativity does not imply $\|I_n\|_*\leq 1$. For all positive powers, induction and submultiplicativity give the estimate $\|X^m\|_*\leq \|X\|_*^m$. Dividing by $m!$ reduces the absolute convergence of the matrix series to the convergence of the scalar exponential series.
[/proofplan]
[step:Separate the zeroth term from the positive powers]
Fix $X\in M_n(\mathbb C)$, and define the scalar
\begin{align*}
a:=\|X\|_*.
\end{align*}
Since $X^0=I_n$, the zeroth term of the series of norms is
\begin{align*}
\left\|\frac{X^0}{0!}\right\|_*=\|I_n\|_*.
\end{align*}
It remains to estimate the terms with $m\geq 1$.
[/step]
[step:Bound every positive matrix power by submultiplicativity]
We prove by induction on $m\in\mathbb N$ that
\begin{align*}
\|X^m\|_* \leq a^m.
\end{align*}
For $m=1$, this is the identity $\|X\|_*=a$. Suppose the estimate holds for some $m\in\mathbb N$. Since the norm is submultiplicative,
\begin{align*}
\|X^{m+1}\|_*=\|X^mX\|_* \leq \|X^m\|_*\|X\|_*.
\end{align*}
Using the induction hypothesis and the definition of $a$, we get
\begin{align*}
\|X^{m+1}\|_* \leq a^m a=a^{m+1}.
\end{align*}
Thus the estimate holds for every $m\geq 1$.
[guided]
The point of this step is to turn a matrix-valued problem into a scalar estimate. Define
\begin{align*}
a:=\|X\|_*.
\end{align*}
We claim that for every positive integer $m$,
\begin{align*}
\|X^m\|_* \leq a^m.
\end{align*}
We prove this by induction. For $m=1$, the statement reads
\begin{align*}
\|X\|_* \leq a,
\end{align*}
which is equality by the definition of $a$. Now assume that the estimate has already been proved for a fixed $m\in\mathbb N$, so
\begin{align*}
\|X^m\|_* \leq a^m.
\end{align*}
Because $\|\cdot\|_*$ is submultiplicative and $X^{m+1}=X^mX$, we have
\begin{align*}
\|X^{m+1}\|_*=\|X^mX\|_* \leq \|X^m\|_*\|X\|_*.
\end{align*}
Substituting the induction hypothesis and $\|X\|_*=a$ gives
\begin{align*}
\|X^{m+1}\|_* \leq a^m a=a^{m+1}.
\end{align*}
This proves the desired estimate for all $m\geq 1$. Notice that we did not use this estimate at $m=0$, because it would require $\|I_n\|_*\leq 1$, which is not part of the hypotheses.
[/guided]
[/step]
[step:Compare the positive tail with the scalar exponential series]
For every $m\geq 1$, homogeneity of the norm gives
\begin{align*}
\left\|\frac{X^m}{m!}\right\|_*=\frac{\|X^m\|_*}{m!}.
\end{align*}
Using the power estimate from the previous step,
\begin{align*}
\left\|\frac{X^m}{m!}\right\|_* \leq \frac{a^m}{m!}.
\end{align*}
The scalar series $\sum_{m=0}^{\infty}a^m/m!$ converges by the [ratio test](/theorems/174): if $a=0$, all terms with $m\geq 1$ vanish; if $a>0$, then for the positive scalar sequence $c_m:=a^m/m!$,
\begin{align*}
\frac{c_{m+1}}{c_m}=\frac{a}{m+1}\to 0.
\end{align*}
Therefore
\begin{align*}
\sum_{m=1}^{\infty}\left\|\frac{X^m}{m!}\right\|_* \leq \sum_{m=1}^{\infty}\frac{a^m}{m!}=e^a-1.
\end{align*}
[/step]
[step:Combine the zeroth term with the positive tail]
Adding the zeroth term computed above to the positive-tail estimate gives
\begin{align*}
\sum_{m=0}^{\infty}\left\|\frac{X^m}{m!}\right\|_* \leq \|I_n\|_*+\sum_{m=1}^{\infty}\frac{a^m}{m!}.
\end{align*}
Since $a=\|X\|_*$, this is
\begin{align*}
\sum_{m=0}^{\infty}\left\|\frac{X^m}{m!}\right\|_* \leq \|I_n\|_*+e^{\|X\|_*}-1<\infty.
\end{align*}
By definition, finiteness of this sum of norms is absolute convergence of the matrix exponential series with respect to $\|\cdot\|_*$. This proves the theorem.
[/step]
