[proofplan]
We use the power-series definition of the matrix exponential. Absolute convergence, supplied by [citetheorem:8777], allows us to multiply the two exponential series by the Cauchy product and regroup terms by total degree. Since both matrices are scalar multiples of the same matrix $X$, their powers commute and the coefficient of $X^k$ is the usual [binomial coefficient](/page/Binomial%20Coefficient) sum. The inverse formula follows by applying the exponential law to the parameters $t$ and $-t$.
[/proofplan]
[step:Expand both exponentials as absolutely convergent matrix series]
Fix $s,t \in \mathbb C$. Let $\|\cdot\|$ be any submultiplicative matrix norm on $M(n,\mathbb C)$. By [citetheorem:8777], the matrix exponential is defined by the absolutely convergent series
\begin{align*}
\exp(A)=\sum_{m=0}^{\infty}\frac{A^m}{m!}
\end{align*}
for every $A \in M(n,\mathbb C)$. Applying this to $A=sX$ and $A=tX$ gives
\begin{align*}
\exp(sX)=\sum_{m=0}^{\infty}\frac{(sX)^m}{m!}
\end{align*}
and
\begin{align*}
\exp(tX)=\sum_{\ell=0}^{\infty}\frac{(tX)^\ell}{\ell!}.
\end{align*}
[/step]
[step:Multiply the absolutely convergent series by the Cauchy product]
Define sequences $(A_m)_{m\geq 0}$ and $(B_\ell)_{\ell\geq 0}$ in $M(n,\mathbb C)$ by
\begin{align*}
A_m=\frac{(sX)^m}{m!}
\end{align*}
and
\begin{align*}
B_\ell=\frac{(tX)^\ell}{\ell!}.
\end{align*}
The two series $\sum_{m=0}^{\infty}A_m$ and $\sum_{\ell=0}^{\infty}B_\ell$ are absolutely convergent. Since the norm is submultiplicative,
\begin{align*}
\sum_{m=0}^{\infty}\sum_{\ell=0}^{\infty}\|A_mB_\ell\|\leq \left(\sum_{m=0}^{\infty}\|A_m\|\right)\left(\sum_{\ell=0}^{\infty}\|B_\ell\|\right)<\infty.
\end{align*}
Hence the double series $\sum_{m,\ell\geq 0}A_mB_\ell$ is absolutely convergent, so it may be regrouped by the total degree $k=m+\ell$. Therefore
\begin{align*}
\exp(sX)\exp(tX)=\sum_{k=0}^{\infty}\sum_{m=0}^{k}\frac{(sX)^m(tX)^{k-m}}{m!(k-m)!}.
\end{align*}
[guided]
We need to justify the passage from the product of two infinite matrix series to a single series indexed by total degree. This is the only convergence point in the proof.
Define
\begin{align*}
A_m=\frac{(sX)^m}{m!}
\end{align*}
and
\begin{align*}
B_\ell=\frac{(tX)^\ell}{\ell!}.
\end{align*}
By [citetheorem:8777], both $\sum_{m=0}^{\infty}A_m$ and $\sum_{\ell=0}^{\infty}B_\ell$ converge absolutely in the matrix norm $\|\cdot\|$. The submultiplicativity hypothesis on the norm gives $\|A_mB_\ell\|\leq \|A_m\|\|B_\ell\|$ for all $m,\ell\geq 0$. Thus
\begin{align*}
\sum_{m=0}^{\infty}\sum_{\ell=0}^{\infty}\|A_mB_\ell\|\leq \left(\sum_{m=0}^{\infty}\|A_m\|\right)\left(\sum_{\ell=0}^{\infty}\|B_\ell\|\right)<\infty.
\end{align*}
So the double series of products is absolutely convergent. Absolute convergence is exactly what permits the Cauchy product regrouping: the product of the two sums equals the sum over all pairs $(m,\ell)$, and this absolutely convergent double sum may be collected along the finite diagonals $m+\ell=k$. Therefore
\begin{align*}
\exp(sX)\exp(tX)=\sum_{k=0}^{\infty}\sum_{m=0}^{k}\frac{(sX)^m(tX)^{k-m}}{m!(k-m)!}.
\end{align*}
[/guided]
[/step]
[step:Use commutativity of scalar multiples of one matrix to identify the coefficients]
For each $k\geq 0$ and each $m\in\{0,\dots,k\}$, scalar multiplication commutes with matrix multiplication and powers of the same matrix multiply by adding exponents. Hence
\begin{align*}
(sX)^m(tX)^{k-m}=s^m t^{k-m}X^mX^{k-m}=s^m t^{k-m}X^k.
\end{align*}
Substituting this into the Cauchy product gives
\begin{align*}
\exp(sX)\exp(tX)=\sum_{k=0}^{\infty}\left(\sum_{m=0}^{k}\frac{s^m t^{k-m}}{m!(k-m)!}\right)X^k.
\end{align*}
For each $k\geq 0$, the binomial identity gives
\begin{align*}
\sum_{m=0}^{k}\frac{s^m t^{k-m}}{m!(k-m)!}=\frac{(s+t)^k}{k!}.
\end{align*}
Therefore
\begin{align*}
\exp(sX)\exp(tX)=\sum_{k=0}^{\infty}\frac{((s+t)X)^k}{k!}=\exp((s+t)X).
\end{align*}
[/step]
[step:Apply the exponential law with opposite parameters to obtain the inverse]
Let $t\in\mathbb C$. Applying the identity already proved with parameters $t$ and $-t$ yields
\begin{align*}
\exp(tX)\exp(-tX)=\exp(0X)=I.
\end{align*}
Applying the same identity with parameters $-t$ and $t$ yields
\begin{align*}
\exp(-tX)\exp(tX)=\exp(0X)=I.
\end{align*}
Thus $\exp(-tX)$ is both a right inverse and a left inverse for $\exp(tX)$. Hence $\exp(tX)$ is invertible, so $\exp(tX)\in GL(n,\mathbb C)$, and its inverse is
\begin{align*}
\exp(tX)^{-1}=\exp(-tX).
\end{align*}
This proves both assertions.
[/step]
