I'd like to show that, given to hermitian operators $A,G$ on a Hilbert space $\mathscr{H}$, the following identity holds:
$$
e^{iG\lambda}A e^{-iG\lambda} = A + i\lambda [G,A] + \frac{\left(i\lambda\right)^2}{2!}[G,[G,A]]+\ldots+\frac{(i\lambda)^n}{n!}\underbrace{[G,[G,[G,\ldots[G}_{n\ times},A]]]\ldots]+\ldots
$$
where $\lambda$ denotes a real parameter and $[\ \!,]$ indicates the commutator.
This is a proof left to the reader by Sakurai in his books on Modern Quantum Mechanics.
Best Answer
Using the series definition of exponential:
$$ e^{iG\lambda}A e^{-iG\lambda} = \sum_{p=0}^\infty\frac{(iG\lambda)^p}{p!}A\sum_{q=0}^\infty\frac{(-iG\lambda)^q}{q!} = \sum_{p=0}^\infty\sum_{q=0}^\infty(-)^q\frac{(i\lambda)^{p+q}}{p!q!}G^pAG^q=\\ \sum_{s=0}^\infty\sum_{d=0}^s(-)^d\frac{(i\lambda)^s}{d!(s-d)!}G^{s-d}AG^d=\\ A+i\lambda[G,A]+\frac{(i\lambda)^2}{2!}[G,[G,A]]+\ldots+\frac{(i\lambda)^n}{n!}\sum_{k=0}^n(-)^k \binom{n}{k}G^{n-k}AG^k+\ldots $$ So we are left with the following relation which we have to verify, and which would prove the statement: $$ \mathscr{F}(n): \sum_{k=0}^n(-)^k \binom{n}{k}G^{n-k}AG^k=\underbrace{[G,[G,[G,\ldots[G}_{n\ times},A]]]\ldots]. $$ Proceeding by induction, since the first terms shown above are compatible with the formula, we have to show that, if $\mathscr{F}$(n) holds then $\mathscr{F}$(n+1) is true as well.
To do this we exploit: $$ \underbrace{[G,[G,[G,\ldots[G}_{n+1\ times},A]]]]\ldots] = \underbrace{[G,[G,[G,\ldots[G}_{n\ times},[G,A]]]\ldots] $$
Then substituting $\mathscr{F}(n)$ yields: $$ \underbrace{[G,[G,[G,\ldots[G}_{n+1\ times},A]]]]\ldots]= \sum_{k=0}^n(-)^k \binom{n}{k}G^{n-k}(GA-AG)G^k =\\ \sum_{k=0}^n(-)^k \binom{n}{k}G^{n+1-k}AG^{k}-\sum_{k=0}^n(-)^k \binom{n}{k}G^{n-k}AG^{k+1}=\\ G^{n+1}A+\sum_{k=1}^n(-)^k \binom{n}{k}G^{n+1-k}AG^{k}-\sum_{k'=1}^{n}(-)^{k'-1} \binom{n}{k'-1}G^{n+1-k'}AG^{k'}+(-)^{n+1}AG^{n+1} $$ where in the last passage we changed summing index in the second sum, and took out the first term from the first and the last from the second. Now: $$ \binom{n}{k}+\binom{n}{k-1} = \binom{n+1}{k} $$ which gives $$ \ldots=G^{n+1}A + \sum_{k=1}^n(-)^k \binom{n+1}{k}G^{n+1-k}AG^{k} + (-)^{n+1}AG^{n+1}= \sum_{k=0}^{n+1}(-)^k \binom{n+1}{k}G^{n+1-k}AG^{k}.$$
And therefore $\mathscr{F}$(n+1) holds.