[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 $m\geq 1$.[/step]

[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: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]