[proofplan]
We first use the exponential law for one fixed matrix to show that $t\mapsto \exp(tX)$ is a homomorphism from $(\mathbb R,+)$ to $GL(n,\mathbb C)$. Its smoothness and derivative at zero follow directly from the absolutely convergent [power series](/page/Power%20Series) for the matrix exponential. For uniqueness, an arbitrary smooth one-parameter subgroup $\gamma$ with $\gamma'(0)=X$ is shown to satisfy the linear matrix differential equation $A'(t)=A(t)X$ with initial condition $A(0)=I_n$. Since $t\mapsto \exp(tX)$ satisfies the same [initial value problem](/page/Initial%20Value%20Problem), uniqueness for finite-dimensional linear ordinary differential equations gives equality for all real $t$.
[/proofplan]
[step:Use the exponential law to obtain a smooth one-parameter subgroup]
Fix $X\in M(n,\mathbb C)$. Define
\begin{align*}
\gamma_X:\mathbb R \to GL(n,\mathbb C),\qquad t\mapsto \exp(tX).
\end{align*}
By [Exponential Law for One Matrix][citetheorem:8778], for every $s,t\in\mathbb R$,
\begin{align*}
\exp((s+t)X)=\exp(sX)\exp(tX).
\end{align*}
The same result also gives $\exp(tX)\in GL(n,\mathbb C)$, with inverse $\exp(-tX)$. Hence $\gamma_X$ is a [group homomorphism](/page/Group%20Homomorphism) from $(\mathbb R,+)$ to $GL(n,\mathbb C)$.
The matrix exponential is defined by an absolutely convergent power series in the finite-dimensional [vector space](/page/Vector%20Space) $M(n,\mathbb C)$:
\begin{align*}
\exp(tX)=\sum_{m=0}^{\infty}\frac{t^mX^m}{m!}.
\end{align*}
[Absolute convergence of the matrix exponential series](/theorems/8777), as in [Absolute Convergence of the Matrix Exponential Series][citetheorem:8777], gives [uniform convergence](/page/Uniform%20Convergence) on compact intervals after applying it to the bounded family $tX$ with $t$ in a compact interval. More generally, fix an integer $r\geq 0$. The formally $r$-times differentiated series is
\begin{align*}
\sum_{m=r}^{\infty}\frac{t^{m-r}X^m}{(m-r)!}.
\end{align*}
On every compact interval this series converges uniformly by the same submultiplicative norm estimate, because it is bounded by the scalar exponential series multiplied by $\|X\|_*^r$. Hence the standard termwise differentiation theorem for uniformly convergent derivative series applies inductively for every $r\geq 0$. Therefore $\gamma_X$ is $C^\infty$ as a map into the finite-dimensional real vector space $M(n,\mathbb C)$, and its first derivative is
\begin{align*}
\gamma_X'(t)=\sum_{m=1}^{\infty}\frac{t^{m-1}X^m}{(m-1)!}.
\end{align*}
Evaluating at $t=0$ gives
\begin{align*}
\gamma_X'(0)=X.
\end{align*}
Thus $\gamma_X$ is a smooth one-parameter subgroup with initial tangent vector $X$.
[guided]
We need two facts: the curve must land in $GL(n,\mathbb C)$ and it must respect addition in the parameter. Define
\begin{align*}
\gamma_X:\mathbb R \to GL(n,\mathbb C),\qquad t\mapsto \exp(tX).
\end{align*}
The target is legitimate because [Exponential Law for One Matrix][citetheorem:8778] implies that $\exp(tX)$ is invertible, with inverse $\exp(-tX)$, for every $t\in\mathbb R$. The same theorem applies to the two scalar parameters $s,t\in\mathbb R\subset\mathbb C$ and gives
\begin{align*}
\gamma_X(s+t)=\exp((s+t)X)=\exp(sX)\exp(tX)=\gamma_X(s)\gamma_X(t).
\end{align*}
This is exactly the homomorphism property for a one-parameter subgroup.
It remains in this step to justify smoothness and compute the initial derivative. The exponential is given by the power series
\begin{align*}
\exp(tX)=\sum_{m=0}^{\infty}\frac{t^mX^m}{m!}.
\end{align*}
Choose any submultiplicative matrix norm $\|\cdot\|_*$ on $M(n,\mathbb C)$ and any compact interval $K\subset\mathbb R$. If $R:=\sup\{|t|:t\in K\}$, then
\begin{align*}
\left\|\frac{t^mX^m}{m!}\right\|_* \leq \frac{R^m\|X\|_*^m}{m!}
\end{align*}
for all $t\in K$. The scalar majorant has finite sum, so the Weierstrass test gives uniform convergence of the exponential series on $K$.
To prove smoothness, not merely differentiability once, we repeat this estimate for every derivative order. Fix an integer $r\geq 0$. The formally $r$-times differentiated series is
\begin{align*}
\sum_{m=r}^{\infty}\frac{t^{m-r}X^m}{(m-r)!}.
\end{align*}
For $t\in K$ and $m\geq r$, submultiplicativity gives
\begin{align*}
\left\|\frac{t^{m-r}X^m}{(m-r)!}\right\|_* \leq \|X\|_*^r\frac{R^{m-r}\|X\|_*^{m-r}}{(m-r)!}.
\end{align*}
The scalar majorant has finite sum, equal to $\|X\|_*^r\exp(R\|X\|_*)$. Hence, for each fixed $r$, the $r$-th formally differentiated series converges uniformly on $K$. The usual termwise differentiation theorem for series of functions into a finite-dimensional [normed vector space](/page/Normed%20Vector%20Space) therefore applies inductively: the zeroth series is continuous, the first derivative series is uniformly convergent on compact intervals, and the same argument repeats at every order. Thus $\gamma_X$ is $C^\infty$ as a map $\mathbb R\to M(n,\mathbb C)$.
For the first derivative, the preceding formula with $r=1$ gives
\begin{align*}
\gamma_X'(t)=\sum_{m=1}^{\infty}\frac{t^{m-1}X^m}{(m-1)!}.
\end{align*}
At $t=0$, every term with $m\geq 2$ contains the factor $0^{m-1}$, while the term $m=1$ is $X$. Thus
\begin{align*}
\gamma_X'(0)=X.
\end{align*}
[/guided]
[/step]
[step:Differentiate an arbitrary one-parameter subgroup to get a linear matrix ODE]
Let
\begin{align*}
\gamma:\mathbb R\to GL(n,\mathbb C)
\end{align*}
be a smooth one-parameter subgroup satisfying $\gamma'(0)=X$. Since $\gamma$ is a group homomorphism from $(\mathbb R,+)$ to $GL(n,\mathbb C)$,
\begin{align*}
\gamma(t+h)=\gamma(t)\gamma(h)
\end{align*}
for all $t,h\in\mathbb R$. Also $\gamma(0)=I_n$, because $\gamma(0)=\gamma(0+0)=\gamma(0)^2$ and $\gamma(0)$ is invertible.
Fix $t\in\mathbb R$. Define the translated curve
\begin{align*}
\eta_t:\mathbb R\to M(n,\mathbb C),\qquad h\mapsto \gamma(t+h).
\end{align*}
Define also
\begin{align*}
\mu_t:\mathbb R\to M(n,\mathbb C),\qquad h\mapsto \gamma(t)\gamma(h).
\end{align*}
The equality $\eta_t=\mu_t$ holds for all $h$. Differentiating at $h=0$ in the ambient vector space $M(n,\mathbb C)$ gives
\begin{align*}
\eta_t'(0)=\mu_t'(0).
\end{align*}
By the chain rule, $\eta_t'(0)=\gamma'(t)$. Since left multiplication by the fixed matrix $\gamma(t)$ is a complex-[linear map](/page/Linear%20Map) on $M(n,\mathbb C)$, differentiating $\mu_t$ gives
\begin{align*}
\mu_t'(0)=\gamma(t)\gamma'(0)=\gamma(t)X.
\end{align*}
Therefore $\gamma$ satisfies the differential equation
\begin{align*}
\gamma'(t)=\gamma(t)X
\end{align*}
for every $t\in\mathbb R$, with initial condition $\gamma(0)=I_n$.
[/step]
[step:Verify that the exponential curve satisfies the same initial value problem]
From the termwise derivative computation in the first step,
\begin{align*}
\gamma_X'(t)=\sum_{m=1}^{\infty}\frac{t^{m-1}X^m}{(m-1)!}.
\end{align*}
Reindexing the sum by $k=m-1$ gives
\begin{align*}
\gamma_X'(t)=\sum_{k=0}^{\infty}\frac{t^kX^{k+1}}{k!}.
\end{align*}
Because $X^{k+1}=X^kX$ for every $k\geq 0$, this becomes
\begin{align*}
\gamma_X'(t)=\left(\sum_{k=0}^{\infty}\frac{t^kX^k}{k!}\right)X=\gamma_X(t)X.
\end{align*}
Also
\begin{align*}
\gamma_X(0)=\exp(0X)=I_n.
\end{align*}
Thus $\gamma_X$ satisfies the same initial value problem
\begin{align*}
A'(t)=A(t)X,\qquad A(0)=I_n.
\end{align*}
[/step]
[step:Apply uniqueness for finite-dimensional linear ODEs]
View $M(n,\mathbb C)$ as a finite-dimensional real vector space. Define the linear map
\begin{align*}
L:M(n,\mathbb C)\to M(n,\mathbb C),\qquad A\mapsto AX.
\end{align*}
The curves $\gamma$ and $\gamma_X$ are smooth maps from $\mathbb R$ into this finite-dimensional real vector space, and the previous two steps show that both solve
\begin{align*}
A'(t)=L(A(t)),\qquad A(0)=I_n.
\end{align*}
We use the uniqueness part of the Picard--Lindelof theorem for ordinary differential equations in finite-dimensional real vector spaces: a continuously differentiable vector field that is locally Lipschitz in the state variable has at most one solution through a prescribed initial value on any common interval of definition. Its hypotheses hold here because $L$ is a linear map on the finite-dimensional real normed vector space $M(n,\mathbb C)$, hence continuous and globally Lipschitz. Therefore the two solutions agree on their common interval of definition $\mathbb R$. Hence
\begin{align*}
\gamma(t)=\gamma_X(t)=\exp(tX)
\end{align*}
for every $t\in\mathbb R$. This proves uniqueness and completes the proof.
[/step]
