Let $X,Y$ be in the lie algebra of a lie group, I want to show that the following identity is true.
$$[X,Y]= \frac{d}{ds}\Big|_{s=0}\frac{d}{dt}\Big|_{t=0}\exp(sX)\exp(tY)\exp(-sX)\exp(-tY).$$
I'm getting stuck and I feel some of the steps I used may not be allowed. Here is my attempt $$\frac{d}{ds}\Big|_{s=0}\frac{d}{dt}\Big|_{t=0}\exp(sX)\exp(tY)\exp(-sX)\exp(-tY)$$ using $a \exp(V) a^{-1}=exp(\text{Ad}(a)V)$ $$=\frac{d}{ds}\Big|_{s=0}\frac{d}{dt}\Big|_{t=0}\exp(sX)\exp(-s \text{Ad}(\exp(tY))X)$$ $$=\frac{d}{ds}\Big|_{s=0}\exp(sX)\frac{d}{dt}\Big|_{t=0}\exp(-s \text{Ad}(\exp(tY))X)$$ $$=\frac{d}{ds}\Big|_{s=0}\exp(sX)\left(\frac{d}{dt}\Big|_{t=0}\left(-s \text{Ad}(\exp(tY))X\right)\right)\exp(-s \text{Ad}(\exp(tY))X)$$
using $ \text{ad}(X)=\frac{d}{dt}\Big|_{t=0} \text{Ad}(\exp(tX)) $
$$=\frac{d}{ds}\Big|_{s=0}\exp(sX)\left(\left(-s \text{ad}(Y)X\right)\right)\exp(-s \text{Ad}(\exp(tY))X)$$
I'm not sure where to go from I tried working out the product rule but that did not seem to help. I also tried to work backwards from the definition of $[X,Y]$ to try and guide me
$$[X,Y]=\text(ad)(X)Y=\frac{d}{dt}\Big|_{t=0}\text{Ad}(\exp(tX))Y$$ but this also did not help
