[proofplan] We first interpret the matrix exponential by its [power series](/page/Power%20Series) and show that the curve $t\mapsto \exp(tX)$ is smooth as a map into the ambient real [vector space](/page/Vector%20Space) $M(n,\mathbb F)$. The exponential law for one matrix gives both invertibility, with inverse $\exp(-tX)$, and the identity $\exp((s+t)X)=\exp(sX)\exp(tX)$. Since $GL(n,\mathbb F)$ is an open submanifold of $M(n,\mathbb F)$, ambient smoothness together with invertibility proves that $\gamma_X$ is a smooth curve in $GL(n,\mathbb F)$ and the exponential law proves the homomorphism property. [/proofplan] [step:Define the exponential curve by an absolutely convergent matrix series] Fix $X\in M(n,\mathbb F)$. For each $t\in\mathbb R$, define \begin{align*} \gamma_X(t)=\exp(tX)=\sum_{m=0}^{\infty}\frac{t^mX^m}{m!}. \end{align*} If $\mathbb F=\mathbb R$, then every partial sum lies in $M(n,\mathbb R)$, so the limit lies in $M(n,\mathbb R)$. If $\mathbb F=\mathbb C$, the same series is taken in the complex vector space $M(n,\mathbb C)$, viewed as a real vector space for smoothness. The absolute convergence of this series follows from [citetheorem:8777] applied to the matrix $tX\in M(n,\mathbb C)$. [/step] [step:Differentiate the power series uniformly on compact intervals] Let $\|\cdot\|$ be a submultiplicative matrix norm on $M(n,\mathbb F)$, regarded as a real norm when $\mathbb F=\mathbb C$. Fix a compact interval $K\subset\mathbb R$, and define \begin{align*} R_K=\sup_{t\in K}|t|. \end{align*} For every integer $k\ge 0$, the $k$th termwise derivative of the series is \begin{align*} \frac{d^k}{dt^k}\left(\frac{t^mX^m}{m!}\right)=\frac{t^{m-k}X^m}{(m-k)!} \end{align*} for $m\ge k$, and is $0$ for $m<k$. For $t\in K$ and $m\ge k$, submultiplicativity gives \begin{align*} \left\|\frac{t^{m-k}X^m}{(m-k)!}\right\|\le \|X\|^k\frac{(R_K\|X\|)^{m-k}}{(m-k)!}. \end{align*} The majorant series \begin{align*} \sum_{m=k}^{\infty}\|X\|^k\frac{(R_K\|X\|)^{m-k}}{(m-k)!}=\|X\|^k e^{R_K\|X\|} \end{align*} is finite. Hence the original power series and all of its termwise derivative series converge uniformly on $K$ by the Weierstrass test. By the standard theorem on termwise differentiation for series of smooth functions with locally uniformly convergent derivative series of every order, the sum is $C^{\infty}$ on $K$. Since $K\subset\mathbb R$ was arbitrary, $\gamma_X:\mathbb R\to M(n,\mathbb F)$ is smooth as a real-valued smooth map into the finite-dimensional real vector space $M(n,\mathbb F)$. In particular, the first derivative is \begin{align*} \gamma_X'(t)=\sum_{m=1}^{\infty}\frac{t^{m-1}X^m}{(m-1)!}=X\sum_{m=0}^{\infty}\frac{t^mX^m}{m!}=X\exp(tX). \end{align*} Since $X$ commutes with each power $X^m$, the same computation also gives \begin{align*} \gamma_X'(t)=\exp(tX)X. \end{align*} [guided] The point of this step is to justify smoothness from the power series without relying on a finite polynomial expression. We choose a submultiplicative matrix norm $\|\cdot\|$ on $M(n,\mathbb F)$; such a norm lets us estimate powers by $\|X^m\|\le \|X\|^m$. When $\mathbb F=\mathbb C$, this is still a norm on the underlying real vector space, which is the relevant structure because $GL(n,\mathbb C)$ is being viewed as a real Lie group. Fix a compact interval $K\subset\mathbb R$ and define \begin{align*} R_K=\sup_{t\in K}|t|. \end{align*} For a fixed derivative order $k\ge 0$, differentiating the $m$th summand gives \begin{align*} \frac{d^k}{dt^k}\left(\frac{t^mX^m}{m!}\right)=\frac{t^{m-k}X^m}{(m-k)!} \end{align*} when $m\ge k$, and gives $0$ when $m<k$. Thus, for every $t\in K$ and every $m\ge k$, \begin{align*} \left\|\frac{t^{m-k}X^m}{(m-k)!}\right\|\le \frac{R_K^{m-k}\|X\|^m}{(m-k)!}=\|X\|^k\frac{(R_K\|X\|)^{m-k}}{(m-k)!}. \end{align*} The right-hand side is the $m$th term of a convergent exponential series: \begin{align*} \sum_{m=k}^{\infty}\|X\|^k\frac{(R_K\|X\|)^{m-k}}{(m-k)!}=\|X\|^k e^{R_K\|X\|}. \end{align*} By the Weierstrass test, the derivative series of every order converges uniformly on $K$. The standard theorem on termwise differentiation for series of smooth finite-dimensional-valued functions applies because, on this compact interval, the original series and each termwise derivative series converge uniformly. Therefore the sum is $C^{\infty}$ on $K$. Since $K$ was an arbitrary compact interval in $\mathbb R$, this proves that the resulting map $\gamma_X:\mathbb R\to M(n,\mathbb F)$ is smooth as a real smooth map. For the first derivative, the [uniform convergence](/page/Uniform%20Convergence) of the differentiated series gives \begin{align*} \gamma_X'(t)=\sum_{m=1}^{\infty}\frac{t^{m-1}X^m}{(m-1)!}. \end{align*} Factoring one copy of $X$ on the left gives \begin{align*} \gamma_X'(t)=X\sum_{m=0}^{\infty}\frac{t^mX^m}{m!}=X\exp(tX). \end{align*} Factoring the same copy on the right is also valid because every term is a power of the same matrix $X$, so all these powers commute. Hence \begin{align*} \gamma_X'(t)=\exp(tX)X. \end{align*} [/guided] [/step] [step:Show that the curve takes values in the general linear group] Let $I\in M(n,\mathbb F)$ denote the identity matrix. For every $t\in\mathbb R$, the matrices $tX$ and $-tX$ are scalar multiples of the same matrix $X$, so they commute. Thus the exponential law in [citetheorem:8778] applies and gives \begin{align*} \exp(tX)\exp(-tX)=\exp(0X)=I \end{align*} and \begin{align*} \exp(-tX)\exp(tX)=\exp(0X)=I. \end{align*} Thus $\exp(tX)$ is invertible, with inverse $\exp(-tX)$. Therefore $\gamma_X(t)\in GL(n,\mathbb F)$ for every $t\in\mathbb R$. Since $GL(n,\mathbb F)$ has the smooth structure of an open subset of the real vector space $M(n,\mathbb F)$, a map into $GL(n,\mathbb F)$ is smooth if and only if it is smooth as a map into $M(n,\mathbb F)$ after the open inclusion. The preceding step proves this ambient smoothness, so $\gamma_X:\mathbb R\to GL(n,\mathbb F)$ is smooth. [/step] [step:Use the exponential law to prove the homomorphism identity] Let $s,t\in\mathbb R$. The matrices $sX$ and $tX$ commute because they are scalar multiples of $X$. Applying [citetheorem:8778] to the same matrix $X$ gives \begin{align*} \gamma_X(s+t)=\exp((s+t)X)=\exp(sX)\exp(tX)=\gamma_X(s)\gamma_X(t). \end{align*} Also, \begin{align*} \gamma_X(0)=\exp(0X)=I. \end{align*} Hence $\gamma_X$ preserves the group operation from $(\mathbb R,+)$ to $GL(n,\mathbb F)$ and sends the additive identity $0$ to the identity matrix $I$. Therefore $\gamma_X$ is a smooth [group homomorphism](/page/Group%20Homomorphism). [/step]