As we know, Bell number can be given using two formula
$B_N=\sum_{k=0}^{N-1}C_{N-1}^{k}B_k$ (recursive)
$B_N=e^{-1}\sum_{k=0}^{\infty}\frac{k^N}{k!}$ (Dobinski's formula)
Now I want to substitute Dobinski's formula into recursive one to validate it is valid.
Here is what I do
$$ \begin{align} B_N&=\sum_{k=0}^{N-1}C_{N-1}^{k}B_k \\ &=\sum_{k=0}^{N-1}C_{N-1}^{k}(e^{-1}\sum_{i=0}^{\infty}\frac{i^k}{i!}) \\ &=e^{-1}\sum_{i=0}^{\infty}\frac{1}{i!}(\sum_{k=0}^{N-1}C_{N-1}^{k}i^k) \\ &=e^{-1}\sum_{i=0}^{\infty}\frac{(i+1)^{N-1}}{i!} \\ &=e^{-1}\sum_{i=0}^{\infty}\frac{(i+1)^N}{(i+1)!} \\ &=e^{-1}\sum_{k=0}^{\infty}\frac{(k+1)^N}{(k+1)!} \\ ?&=e^{-1}\sum_{k=0}^{\infty}\frac{k^N}{k!} \end{align} $$
But I have no idea how to deal with the line with a question sign? Thank you very much!
Thanks for the clarification given in the comments, using $$j = k + 1$$ and $$\frac{0^N}{0!}=0$$, we have $$ \begin{align} &=e^{-1}\sum_{k=0}^{\infty}\frac{(k+1)^N}{(k+1)!} \\ &=e^{-1}\sum_{j=1}^{\infty}\frac{j^N}{j!} \\ &=e^{-1}(0 + \sum_{j=1}^{\infty}\frac{j^N}{j!}) \\ &=e^{-1}\sum_{k=0}^{\infty}\frac{j^N}{j!} \\ &=e^{-1}\sum_{k=0}^{\infty}\frac{k^N}{k!} \\ &=B_N \end{align} $$