Asume we have 2 noncommutative operators $\hat{A}, \hat{B}$ but they comply this condition $$ [\hat{A},\, [\hat{A},\, \hat{B}]]=[\hat{B},\, [\hat{A},\, \hat{B}]]=0. $$
Find what equals to operator $\hat{C}$ in equation \begin{equation} e^{\hat{A}}e^{\hat{B}}=e^{\hat{A}+\hat{B}}e^{\hat{C}}=e^{\hat{C}}e^{\hat{A}+\hat{B}}=e^{\hat{A}+\hat{B}+\hat{C}} \end{equation} and prove all relations in equation above.
I have tried to assume function $g(s)=e^{s\hat{A}}e^{s\hat{B}}$ and use the BCH formula, which lead me to relation $$e^{s\hat{A}}e^{s\hat{B}}=\exp\left(s(\hat{A}+\hat{B})+\frac{s^2}{2}[\hat{A},\, \hat{B}]+\frac{s^3}{12}([\hat{A},\, [\hat{A},\, \hat{B}]]+[\hat{B},\, [\hat{A},\, \hat{B}]])+O(s^4)\right).$$
I see that terms with $s^3$ and higher orders are zero due to the assumption on commutators in the first equation. I don't know how to continue further.
For $t\in\mathbb{R}$, let $$\hat F(t):=\exp(\hat At)\exp(\hat Bt)\exp\big(-(\hat A+\hat B)t\big)\,.$$ Then, $$\hat F'(t)=\text{e}^{\hat At}\hat A\,\text{e}^{\hat Bt}\,\text{e}^{-(\hat A+\hat B)t}+\text{e}^{\hat At}\,\hat B\text{e}^{\hat Bt}\,\text{e}^{-(\hat A+\hat B)t}-\text{e}^{\hat At}\,\text{e}^{\hat Bt}\,(\hat A+\hat B)\text{e}^{-(\hat A+\hat B)t}\,.$$ Hence, $$\hat F'(t)=\text{e}^{\hat At}\,\Big[(\hat A+\hat B),\text{e}^{\hat Bt}\Big]\,\text{e}^{-(\hat A+\hat B)t}\,.$$ Since $\hat Z:=[\hat A,\hat B]$ commutes with $\hat B$, we obtain $$\left[\hat A,\hat B^k\right]=k\,\hat Z\,\hat B^{k-1}$$ so that $$\left[\hat A,\text{e}^{\hat Bt}\right]=\hat Zt\,\text{e}^{\hat Bt}\,.$$ Because $\hat B$ commutes with $\text{e}^{\hat Bt}$, we get $$\hat F'(t)=\text{e}^{\hat At}\,\hat Zt\,\text{e}^{\hat Bt}\,\text{e}^{-(\hat A+\hat B)t}\,.$$ As $\hat Z$ commutes with $\hat A$ as well, we obtain $$\hat F'(t)=\hat Zt\,\left(\text{e}^{\hat At}\,\text{e}^{\hat Bt}\,\text{e}^{-(\hat A+\hat B)t}\right)=\hat Zt\,\hat F(t)\,.$$ Consequently, $$\hat F(t)=\exp\left(\int_0^t\,\hat Zs\,\text{d}s\right)\,\hat F(0)=\exp\left(\frac{1}{2}\,\hat Zt^2\right)\,,$$ since $\hat F(0)$ is the identity operator. Hence, $\hat F(1)=\exp\left(\dfrac{1}{2}\,\hat Z\right)$, which implies $$\text{e}^{\hat A}\,\text{e}^{\hat B}=\text{e}^{\hat A+\hat B}\,\text{e}^{\frac{1}{2}\hat Z}=\text{e}^{\frac{1}{2}\hat Z}\,\text{e}^{\hat A+\hat B}=\text{e}^{\hat A+\hat B+\frac{1}{2}\hat Z}\,.$$