Stirling Numbers Exponential Generating Function Induction

Question

I was reading the solution to a question written here, and it uses a fact which can be proved by induction.

The question is to show that an EGF for Stirling Numbers of the second kind with fixed $k$, \begin{eqnarray*} S_k(x)=\sum_{n=k}^{\infty} S_{n,k} \frac{x^n}{n!} \end{eqnarray*} has a closed form: \begin{eqnarray*} S_k(x)=\frac{(e^x-1)^k}{k!} \end{eqnarray*} However, I have been trying for a while and I have been unable to figure out the inductive step, especially when it comes to using the given fact $\frac{d}{dx} S_k(x)=kS_k(x)+S_{k-1}(x)$. Any help to proceed with the inductive step?

Answer 1

We just need to solve the ODE $$ \frac{d}{dx}S_{k+1}(x)-(k+1)S_{k+1}(x)=S_k(x). \tag{1} $$ Multiplying both sides of $(1)$ by the integrating factor $e^{-(k+1)x}$, and using the induction hypothesis $S_k(x)=\frac{(e^x-1)^k}{k!}$, we obtain \begin{align} &\frac{d}{dx}\left[e^{-(k+1)x}S_{k+1}(x)\right]=e^{-(k+1)x}\frac{(e^x-1)^k}{k!} =\frac{e^{-x}(1-e^{-x})^k}{k!} \\ &\quad\implies S_{k+1}(x)=e^{(k+1)x}\left(C+\int\frac{e^{-x}(1-e^{-x})^k}{k!} \,dx\right). \tag{2} \end{align} To compute the integral, we make the substitution $u=-e^{-x}, u=e^{-x}dx$: \begin{align} S_{k+1}(x)&=e^{(k+1)x}\left(C+\int\frac{(1+u)^k}{k!}\,du\right) \\ &=e^{(k+1)x}\left(C+\frac{(1+u)^{k+1}}{(k+1)!}\right) \\ &=e^{(k+1)x}\left(C+\frac{(1-e^{-x})^{k+1}}{(k+1)!}\right) \\ &=Ce^{(k+1)x}+\frac{(e^x-1)^{k+1}}{(k+1)!}. \tag{3} \end{align} To determine the constant $C$, we notice that $S_{k+1}(x)=\sum_{n=k+1}^{\infty} S_{n,k+1} \frac{x^n}{n!}$ implies $S_{k+1}(0)=0$, hence $C=0$. Therefore, $$ S_{k+1}(x)=\frac{(e^x-1)^{k+1}}{(k+1)!}.\quad{\square} \tag{4} $$