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?