(1)

\[ e^{\lambda A}Be^{-\lambda A}=\sum_{k=0}^{\infty}\frac{F_{k}\left(A,B\right)}{k!}\lambda^{k} \] とおくと、
\begin{align*} \frac{d}{d\lambda}e^{\lambda A}Be^{-\lambda A} & =Ae^{\lambda A}Be^{-\lambda A}+e^{\lambda A}B\left(-A\right)e^{-\lambda A}\\ & =Ae^{\lambda A}Be^{-\lambda A}-e^{\lambda A}Be^{-\lambda A}A\\ & =\left[A,e^{\lambda A}Be^{-\lambda A}\right]\\ & =\left[A,\sum_{k=0}^{\infty}\frac{F_{k}\left(A,B\right)}{k!}\lambda^{k}\right]\\ & =\sum_{k=0}^{\infty}\frac{\lambda^{k}}{k!}\left[A,F_{k}\left(A,B\right)\right] \end{align*} \begin{align*} \frac{d}{d\lambda}\sum_{k=0}^{\infty}\frac{F_{k}\left(A,B\right)}{k!}\lambda^{k} & =\sum_{k=1}^{\infty}\frac{F_{k}\left(A,B\right)}{\left(k-1\right)!}\lambda^{k-1}\\ & =\sum_{k=0}^{\infty}\frac{F_{k+1}\left(A,B\right)}{k!}\lambda^{k} \end{align*} これを比べて、
\[ F_{k+1}\left(A,B\right)=\left[A,F_{k}\left(A,B\right)\right] \] \(\lambda=0\)のとき、\(B=F_{0}\left(A,B\right)\)となるので、
\begin{align*} e^{\lambda A}Be^{-\lambda A} & =\sum_{k=0}^{\infty}\frac{F_{k}\left(A,B\right)}{k!}\lambda^{k}\\ & =F_{0}\left(A,B\right)+F_{1}\left(A,B\right)\lambda+\frac{1}{2}F_{2}\left(A,B\right)\lambda^{2}+\frac{1}{3!}F_{3}\left(A,B\right)\lambda^{3}+\cdots+\frac{1}{n!}F_{n}\left(A,B\right)\lambda^{n}+\cdots\\ & =B+\left[A,B\right]\lambda+\frac{1}{2}\left[A,\left[A,B\right]\right]\lambda^{2}+\frac{1}{3!}\left[A,\left[A,\left[A,B\right]\right]\right]\lambda^{3}+\cdots+\frac{1}{n!}\left[A,\left[A,\cdots\left[A,B\right]\right]\right]\lambda^{n}+\cdots \end{align*} となり、\(\lambda=1\)を代入すると与式になる。
従って与式は成り立つ。

(2)

\[ f\left(\lambda\right)=e^{\lambda A}e^{\lambda B} \] とおくと、
\begin{align*} f'\left(\lambda\right) & =Ae^{\mathit{\lambda A}}e^{\lambda B}+e^{\lambda A}Be^{\lambda B}\\ & =e^{\mathit{\lambda A}}Ae^{\lambda B}+e^{\lambda A}Be^{\lambda B}\\ & =e^{\mathit{\lambda A}}\left(A+B\right)e^{\lambda B} \end{align*} \begin{align*} f''\left(\lambda\right) & =Ae^{\mathit{\lambda A}}\left(A+B\right)e^{\lambda B}+e^{\mathit{\lambda A}}\left(A+B\right)Be^{\lambda B}\\ & =e^{\mathit{\lambda A}}\left\{ A\left(A+B\right)+\left(A+B\right)B\right\} e^{\lambda B}\\ & =e^{\mathit{\lambda A}}\left(A^{2}+2AB+B^{2}\right)e^{\lambda B}\\ & =e^{\mathit{\lambda A}}\left(\left(A+B\right)^{2}+\left[A,B\right]\right)e^{\lambda B} \end{align*} \begin{align*} f'''\left(\lambda\right) & =Ae^{\mathit{\lambda A}}\left(A^{2}+2AB+B^{2}\right)e^{\lambda B}+e^{\mathit{\lambda A}}\left(A^{2}+2AB+B^{2}\right)Be^{\lambda B}\\ & =e^{\mathit{\lambda A}}\left\{ A\left(A^{2}+2AB+B^{2}\right)+\left(A^{2}+2AB+B^{2}\right)B\right\} e^{\lambda B}\\ & =e^{\mathit{\lambda A}}\left(A^{3}+3A^{2}B+3AB^{2}+B^{3}\right)e^{\lambda B}\\ & =e^{\mathit{\lambda A}}\left\{ \left(A+B\right)^{3}-BA^{2}-ABA+2A^{2}B-B^{2}A-BAB+2AB^{2}\right\} e^{\lambda B}\\ & =e^{\mathit{\lambda A}}\left\{ \left(A+B\right)^{3}+\left[A^{2},B\right]+A\left[A,B\right]+\left[A,B^{2}\right]+\left[A,B\right]B\right\} e^{\lambda B} \end{align*} となるので、\(\lambda=0\)を代入すると、
\[ f'\left(0\right)M=A+B \] \[ f''\left(0\right)=\left(A+B\right)^{2}+\left[A,B\right] \] \begin{align*} f'''\left(0\right) & =\left(A+B\right)^{3}+\left[A^{2},B\right]+A\left[A,B\right]+\left[A,B^{2}\right]+\left[A,B\right]B \end{align*} となる。
これより、
\begin{align*} f\left(\lambda\right) & =\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{k!}\lambda^{k}\\ & =\exp\left(\log\left(\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{k!}\lambda^{k}\right)\right)\\ & =\exp\left(\log\left(1+\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{k!}\lambda^{k}\right)\right)\\ & =\exp\left(\sum_{j=1}^{\infty}\frac{\left(-1\right)^{j+1}}{j}\left(\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{k!}\lambda^{k}\right)^{j}\right)\cmt{\because\log\left(1+x\right)=\sum_{j=1}^{\infty}\frac{\left(-1\right)^{j+1}}{j}x^{j}}\\ & =\exp\left(\sum_{j=1}^{\infty}\frac{\left(-1\right)^{j+1}}{j}\left(f^{\left(1\right)}\left(0\right)\lambda+\frac{1}{2}f^{\left(2\right)}\left(0\right)\lambda^{2}+\frac{1}{3!}f^{\left(3\right)}\left(0\right)\lambda^{3}+\cdots\right)^{j}\right)\\ & =\exp\left(f^{\left(1\right)}\left(0\right)\lambda+\left(\frac{1}{2}f^{\left(2\right)}\left(0\right)-\frac{1}{2}f^{\left(1\right),2}\left(0\right)\right)\lambda^{2}+\left(\frac{1}{3!}f^{\left(3\right)}\left(0\right)-\frac{2}{2^{2}}\left(f^{\left(1\right)}\left(0\right)f^{\left(2\right)}\left(0\right)+f^{\left(2\right)}\left(0\right)f^{\left(1\right)}\left(0\right)\right)+\frac{1}{3}f^{\left(1\right),3}\left(0\right)\right)\lambda^{3}+\cdots\right)\\ & =\exp\left(\left(A+B\right)\lambda+\left(\frac{1}{2}\left(\left(A+B\right)^{2}+\left[A,B\right]\right)-\frac{1}{2}\left(A+B\right)^{2}\right)\lambda^{2}+\left(\frac{1}{3!}\left(\left(A+B\right)^{3}+\left[A^{2},B\right]+A\left[A,B\right]+\left[A,B^{2}\right]+\left[A,B\right]B\right)-\frac{1}{2^{2}}\left\{ \left(A+B\right)\left(\left(A+B\right)^{2}+\left[A,B\right]\right)+\left(\left(A+B\right)^{2}+\left[A,B\right]\right)\left(A+B\right)\right\} +\frac{1}{3}\left(A+B\right)^{3}\right)\lambda^{3}+\cdots\right)\\ & =\exp\left(\left(A+B\right)\lambda+\frac{1}{2}\left[A,B\right]\lambda^{2}+\left(\frac{1}{3!}\left(\left[A^{2},B\right]+A\left[A,B\right]+\left[A,B^{2}\right]+\left[A,B\right]B\right)-\frac{1}{2^{2}}\left(\left(A+B\right)\left[A,B\right]+\left[A,B\right]\left(A+B\right)\right)\right)\lambda^{3}+\cdots\right)\\ & =\exp\left(\left(A+B\right)\lambda+\frac{1}{2}\left[A,B\right]\lambda^{2}+\left(\frac{1}{3!}\left(\left[A,B\right]A+2A\left[A,B\right]+B\left[A,B\right]+2\left[A,B\right]B\right)-\frac{1}{2^{2}}\left(\left(A+B\right)\left[A,B\right]\right)+\left[A,B\right]\left(A+B\right)\right)\lambda^{3}+\cdots\right)\\ & =\exp\left(\left(A+B\right)\lambda+\frac{1}{2}\left[A,B\right]\lambda^{2}+\left(-\frac{1}{12}\left[A,B\right]A+\frac{1}{12}A\left[A,B\right]-\frac{1}{12}B\left[A,B\right]+\frac{1}{12}\left[A,B\right]B\right)\lambda^{3}+\cdots\right)\\ & =\exp\left(\left(A+B\right)\lambda+\frac{1}{2}\left[A,B\right]\lambda^{2}+\frac{1}{12}\left(\left(A-B\right)\left[A,B\right]-\left[A,B\right]\left(A-B\right)\right)\lambda^{3}+\cdots\right)\\ & =\exp\left(\left(A+B\right)\lambda+\frac{1}{2}\left[A,B\right]\lambda^{2}+\frac{1}{12}\left[A-B,\left[A,B\right]\right]\lambda^{3}+\cdots\right) \end{align*} となり、\(\lambda=1\)を代入して、
\begin{align*} e^{A}e^{B} & =f\left(1\right)\\ & =\exp\left(A+B+\frac{1}{2}\left[A,B\right]+\frac{1}{12}\left[A-B,\left[A,B\right]\right]+\cdots\right) \end{align*} となるので与式は成り立つ。