[proofplan]
Fix $Y\in\mathfrak g$ and compare two curves in the finite-dimensional real [vector space](/page/Vector%20Space) $\mathfrak g$: the conjugation curve $t\mapsto \operatorname{Ad}_{\exp(tX)}Y$ and the operator-exponential curve $t\mapsto \exp(t\operatorname{ad}_X)Y$. Both have initial value $Y$. Differentiating the conjugation curve shows that it satisfies the same linear differential equation $Z'=\operatorname{ad}_X(Z)$ as the operator-exponential curve, so uniqueness for finite-dimensional linear ODEs identifies the two curves. Evaluating at $t=1$ and using arbitrariness of $Y$ gives equality of the two linear maps.
[/proofplan]
[step:Define the two curves in the Lie algebra]
Fix $Y\in\mathfrak g$. Let $M(n,\mathbb C)$ denote the complex vector space of complex $n\times n$ matrices, and let $I_n\in M(n,\mathbb C)$ denote the identity matrix. Since $X\in\mathfrak g$, the matrix exponential curve
\begin{align*}
\gamma_X:\mathbb R\to G,\quad t\mapsto \exp(tX)
\end{align*}
lies in $G$ by the definition of the [Lie algebra](/page/Lie%20Algebra) of a matrix Lie group. Define
\begin{align*}
F:\mathbb R\to\mathfrak g,\quad t\mapsto \operatorname{Ad}_{\exp(tX)}Y.
\end{align*}
We now compute the differential definition of $\operatorname{Ad}_g$ in matrix form. Fix $g\in G$, and let
\begin{align*}
\alpha:(-\varepsilon,\varepsilon)\to G
\end{align*}
be a smooth curve with $\alpha(0)=I_n$ and $\alpha'(0)=Y$, where $\varepsilon>0$. Then $C_g\circ\alpha:(-\varepsilon,\varepsilon)\to G$ is the smooth curve $s\mapsto g\alpha(s)g^{-1}$, and differentiating in the ambient real vector space $M(n,\mathbb C)$ gives
\begin{align*}
\operatorname{Ad}_g(Y)=d(C_g)_{I_n}(Y)=\frac{d}{ds}\bigg|_{s=0}g\alpha(s)g^{-1}=gYg^{-1}.
\end{align*}
Applying this with $g=\exp(tX)$ gives
\begin{align*}
F(t)=\exp(tX)Y\exp(-tX).
\end{align*}
The value $F(t)$ lies in $\mathfrak g$ because $\operatorname{Ad}_g:\mathfrak g\to\mathfrak g$ for every $g\in G$.
Let
\begin{align*}
L:\mathfrak g\to\mathfrak g,\quad Z\mapsto \operatorname{ad}_X(Z)=[X,Z]=XZ-ZX.
\end{align*}
Define the comparison curve
\begin{align*}
H:\mathbb R\to\mathfrak g,\quad t\mapsto \exp(tL)Y,
\end{align*}
where
\begin{align*}
\exp(tL)=\sum_{k=0}^{\infty}\frac{t^kL^k}{k!}
\end{align*}
is the operator exponential in the finite-dimensional real vector space $\operatorname{End}_{\mathbb R}(\mathfrak g)$.
[/step]
[step:Differentiate the conjugation curve and obtain the linear ODE]
The matrix exponential satisfies
\begin{align*}
\frac{d}{dt}\exp(tX)=X\exp(tX)=\exp(tX)X
\end{align*}
and, using the same formula with $-X$,
\begin{align*}
\frac{d}{dt}\exp(-tX)=-X\exp(-tX)=-\exp(-tX)X.
\end{align*}
Applying the product rule to
\begin{align*}
F(t)=\exp(tX)Y\exp(-tX)
\end{align*}
gives
\begin{align*}
F'(t)=X\exp(tX)Y\exp(-tX)-\exp(tX)Y\exp(-tX)X.
\end{align*}
Therefore
\begin{align*}
F'(t)=XF(t)-F(t)X=[X,F(t)]=\operatorname{ad}_X(F(t))=L(F(t)).
\end{align*}
Also,
\begin{align*}
F(0)=I_nYI_n=Y.
\end{align*}
[guided]
The goal is to turn the conjugation identity into an [ordinary differential equation](/page/Ordinary%20Differential%20Equation). We start from the declared map
\begin{align*}
F:\mathbb R\to\mathfrak g,\quad t\mapsto \operatorname{Ad}_{\exp(tX)}Y=\exp(tX)Y\exp(-tX).
\end{align*}
This curve is differentiable because the matrix exponential is differentiable as a [power series](/page/Power%20Series) in the finite-dimensional real vector space $M(n,\mathbb C)$ of complex $n\times n$ matrices, and matrix multiplication is bilinear.
The derivative of the first exponential factor is
\begin{align*}
\frac{d}{dt}\exp(tX)=X\exp(tX).
\end{align*}
The derivative of the inverse exponential factor is
\begin{align*}
\frac{d}{dt}\exp(-tX)=-\exp(-tX)X.
\end{align*}
The sign appears because the inner matrix is $-tX$. Applying the product rule to the three-factor product gives
\begin{align*}
F'(t)=X\exp(tX)Y\exp(-tX)+\exp(tX)Y\bigl(-\exp(-tX)X\bigr).
\end{align*}
Rewriting the two terms in terms of $F(t)$ yields
\begin{align*}
F'(t)=XF(t)-F(t)X.
\end{align*}
By definition of the infinitesimal adjoint action,
\begin{align*}
\operatorname{ad}_X(Z)=[X,Z]=XZ-ZX
\end{align*}
for every $Z\in\mathfrak g$. Substituting $Z=F(t)$ gives the differential equation
\begin{align*}
F'(t)=\operatorname{ad}_X(F(t)).
\end{align*}
Finally, evaluating at $t=0$ gives
\begin{align*}
F(0)=\exp(0X)Y\exp(0X)=I_nYI_n=Y.
\end{align*}
Thus $F$ is a solution of the linear [initial value problem](/page/Initial%20Value%20Problem)
\begin{align*}
Z'(t)=\operatorname{ad}_X(Z(t)),\qquad Z(0)=Y.
\end{align*}
[/guided]
[/step]
[step:Verify that the operator exponential curve solves the same ODE]
Since $L=\operatorname{ad}_X$ is a linear endomorphism of the finite-dimensional real vector space $\mathfrak g$, termwise differentiation of the operator exponential series gives
\begin{align*}
\frac{d}{dt}\exp(tL)=L\exp(tL)=\exp(tL)L.
\end{align*}
Hence
\begin{align*}
H'(t)=L\exp(tL)Y=L(H(t)).
\end{align*}
Moreover,
\begin{align*}
H(0)=\exp(0L)Y=Y.
\end{align*}
Thus $F$ and $H$ solve the same linear initial value problem in $\mathfrak g$:
\begin{align*}
Z'(t)=L(Z(t)),\qquad Z(0)=Y.
\end{align*}
[/step]
[step:Use uniqueness of linear ODEs and evaluate at time one]
By the uniqueness theorem for linear ordinary differential equations in finite-dimensional real vector spaces, applied to the real vector space $\mathfrak g$ and the [linear map](/page/Linear%20Map) $L:\mathfrak g\to\mathfrak g$, the two solutions of
\begin{align*}
Z'(t)=L(Z(t)),\quad Z(0)=Y
\end{align*}
are equal. Therefore
\begin{align*}
\operatorname{Ad}_{\exp(tX)}Y=F(t)=H(t)=\exp(t\operatorname{ad}_X)Y
\end{align*}
for every $t\in\mathbb R$.
Setting $t=1$ gives
\begin{align*}
\operatorname{Ad}_{\exp X}Y=\exp(\operatorname{ad}_X)Y.
\end{align*}
Since $Y\in\mathfrak g$ was arbitrary, the two real-linear maps $\mathfrak g\to\mathfrak g$ agree:
\begin{align*}
\operatorname{Ad}_{\exp X}=\exp(\operatorname{ad}_X).
\end{align*}
[/step]
