Solve $\displaystyle S_n(x) = x \frac{\partial }{\partial x}(S_{n-1}(x)) , S_0(x) = e^x, S_n(x) = ?$

Question

$\displaystyle S_n(x) = x \frac{\partial }{\partial x}(S_{n-1}(x)), S_0(x) = e^x$

I would like to know what's the method to solve such recurrences.

This is what I got while trying to find $\displaystyle \sum_{k=1}^{\infty}{\frac{k^n}{k!}} $

All I could think of was comparing coefficients but here it is not a polynomial. I also thought about making patterns but doesn't help.

$\displaystyle S_n(x) + S_{n-1}(x) = \frac{\partial}{\partial x}(xS_{n-1}(x))$

and then if I assume $A_{n-1}(x) = xS_{n-1}(x)$, then recurrence keeps on repeating. That is all I could observe.

Answer 1

This is not an answer since the result is just based on observation.

Out of curiosity, I computed the first terms and, as one coud expect, the result is $$S_n(x)=e^x \,P_n(x)$$ Looking at the coefficients, it became quite clear that $$P_n(x)=\sum _{i=1}^n \mathcal{S}_n^{(i)}\, x^i$$ where appear Stirling numbers of the second kind.