My professor wants me to solve this identity in two ways. Sadly, I could only do one way and haven't figure out how to solve it another way. Here is my way,
Denote $F(t)=e^{t\hat{A}}\hat{B}e^{-t\hat{A}}$
$\frac{dF}{dt}=\hat{A}e^{t\hat{A}}\hat{B}e^{-t\hat{A}}-e^{t\hat{A}}\hat{B}e^{-t\hat{A}}\hat{A}=[\hat{A},F]=ad\hat{A}\cdot F(t)\quad\Rightarrow\frac{d^nF}{dt^n}=\left(ad\hat{A}\right)^n\cdot F(t)$
Using Taylor expansion $f(x)=\Sigma_n\frac{f^{(n)}(a)}{n!}(x-a)^n$ for $a=0$
$\begin{array}{l}F(t)&=F(0)+F'(0)t+F''(0)\frac{t}{2}+\dots+\frac{F^{(n)}(0)t^n}{n!}\\&=\hat{B}+ad\hat{A}\cdot\hat{B}+\dots\\&=e^{ad\hat{A}t}\cdot\hat{B}\end{array}$
Thus, $e^{\hat{A}}\hat{B}e^{-\hat{A}}=F(1)=e^{ad\hat{A}}\cdot\hat{B}$
Could you show me another way prove this identity?
Here's another way: consider the ordinary, linear, constant coefficient differential equation
$\dfrac{dX(t)}{dt} = AX - XA = [A, X] = (ad A) X, \tag{1}$
$X(0) = B. \tag{2}$
(1) has a unique solution satisfying (2), and since $ad A = [A, \cdot]$ is a constant linear map, this solution is
$X(t) = e^{t (ad A)} B; \tag{3}$
we note that
$X(0) = e^{0 (ad A)} B = e^0 B = B, \tag{4}$
verifying that (3) indeed fulfils (2). Now set
$Y(t) = e^{tA} B e^{-tA}; \tag{5}$
we find
$\dfrac{dY(t)}{dt} = Ae^{tA}Be^{-tA} - e^{tA}B e^{-tA} A = AY(t) - Y(t) A = (ad A) Y(t); \tag{6}$
we see that $Y(t)$ also obeys (1) and furthermore
$Y(0) = B. \tag{7}$
Since both $X(t)$ and $Y(t)$ satisfy the same differential equation with the same initial condition, uniqueness of the solution implies $X(t) = Y(t)$ for all $t$, whence
$e^{tA} B e^{-tA} = e^{t (ad A)} B \tag{8}$
holds for all $t \in \Bbb R$. Now set $t = 1$ and you're done!