To expand on my comment, observe that
$ vw^* A^n vw^* = \left( w^* A^n v \right) vw^* $
so in any product of $A$'s and $vw^*$'s with more than one copy of $vw^*$, we can convert the middle part to a scalar and extract it.
Applying this and grouping like terms gives the formula
$\begin{align} (A + vw^*)^n = A^n + \sum_{i=0}^{n-1} A^i v w^* A^{n-1-i} + \sum_{i=0}^{n-2} \sum_{j=0}^{n-2-i} A^i v w^* A^j \left( w^* (A + vw^*)^{n-2-i-j} v \right) \end{align}$
Summing this to get $\exp(A + vw^*)$, the second term yields
$ \sum_{n=1}^{+\infty} \frac{1}{n!} \sum_{i=0}^{n-1} A^i v w^* A^{n-1-i} = \sum_{i=0}^{+\infty} A^i v w^* \sum_{n=i+1}^{+\infty} \frac{1}{n!}A^{n-1-i} $
I'm not particularly inclined to deal with truncated exponentials of $A$. :( The third term also involves truncated exponentials of $A + vw^*$.
The path forward with this idea is not clear. I only see two ideas, and both promise to be irritating:
- Try to come up with a simplified formula for all truncated exponentials, hoping the complicated terms cancel or otherwise collect together or have a nice recursion
- Use combinatorics to further simplify $w^* (A + vw^*)^{n-2-i-j} v$ and hope something nice falls out.