This is kind of a question about Lie groups/algebras, but what is really hiding is some combinatorial work and some linear algebra
In the context of Matrix Lie groups we can define the ad $$ad_x(Y)=[X,Y]=XY-YX$$ and Ad maps $$Ad_A(X)=AXA^{-1}$$
I am trying show by direct calculation that $$e^{ad_X}(Y) = Ad_{e^X}(Y)=e^X Ye^{-X}$$
where $X$ and $Y$ are $n \times n$ matricies. (This is a question from Brian Hall's matrix Lie group book)
I have calculated that
$$(ad_X)^m(Y) = \sum_{k=0}^m \binom{m}{k} X^kY(-X)^{m-k}$$
where $$(ad_X)^m(Y)=[X,\ldots,[X,[X,Y]]\cdots].$$
Thus $$e^{ad_X}(Y) = \sum_{p=0}^\infty \frac{(ad_X)^p(Y)}{p!}$$ and so $$e^{ad_X}(Y) = \sum_{p=0}^\infty \sum_{k=0}^p \binom{p}{k} \frac{1}{p!} X^kY(-X)^{p-k}$$
I'm not really sure how to convert this into something that looks like the series expansion of $e^X Y e^{-X}$. Any tips?
Let $L_X(A)=XA$ and $R_X(A)=AX$ (left and right multiplication operators. Then $ad_X(A)=XA-AX=L_X(A)-R_X(A)=(L_X-R_X)(A)$. Note that the left and right multiplication operators commute with each other (because matrix multiplication is associative).
Therefore, $e^{ad_X}=e^{L_X-R_X}=e^{L_X}e^{-R_X}$ (since $e^{U+V}=e^Ue^V$ when $U$ and $V$ commute). So we find that
$$e^{ad_X}A=e^{L_X}e^{-R_X}A=e^{L_X}\sum\limits_{n=0}^\infty \frac{(-1)^nR_X^n(A)}{n!} = e^{L_X}\sum\limits_{n=0}^\infty \frac{(-1)^nAX^n}{n!}=e^{L_X}Ae^{-X}$$
$$ = \sum\limits_{n=0}^\infty \frac{L_X^n}{n!} Ae^{-X} = \sum\limits_{n=0}^\infty \frac{X^nAe^{-X}}{n!} = e^XAe^{-X} = e^XA(e^X)^{-1}=Ad_{e^X}(A)$$