Leibniz rule for error function

Question

Let $f(x)=erfi(a+x)$ and $g(x)=e^{cx}$ with \begin{align*} f^{(n)}(x)=\frac{2}{\sqrt{\pi}}e^{(a+x)^2}\sum_{m=0}^{n-1}\sum_{j=0}^{m}\frac{\binom{m}{j}(-1)^j(a+x)^{2m-n+1}}{m!}\\ \prod_{p=1}^{n-1}(2m-2j-p+1). \end{align*} and \begin{align*} g^{(n)}(x)=c^n e^{cx} \end{align*} I want to find an expression for $ (f(x).g(x))^{(n)}$. I tried Leibniz rule for differentiation \begin{align*} &(f(x).g(x))^{(n)} =\sum_{k=0}^{n}\binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)\\ &\quad=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{2}{\sqrt{\pi}}e^{(a+x)^2}\sum_{m=0}^{k-1}\sum_{j=0}^{m}\frac{\binom{m}{j}(-1)^j(a+x)^{2m-k+1}}{m!} \prod_{p=1}^{k-1}(2m-2j-p+1)\right)\\ &\times\left(c^{(n-k)} e^{cx}\right) \end{align*} I have no other way to find $ f^{(n)}$, the reason is that we need to find first derivative of error function to get rid of integral and then generalize derivative. It should be $ f^{(0)}=f $ for applying Leibniz rule while $f^{(0)}=0 $ in this case. How to overcome this problem?