Does this sequence have a closed form representation?

Question

We know that

$$ \sum_{s=0}^\infty \frac{\lambda^{s}}{s!} = e^\lambda$$

Relatedly,

$$ \sum_{s=1}^\infty \frac{\lambda^{s}}{s!}s = \lambda \sum_{s=1}^\infty \frac{\lambda^{s-1}}{(s-1)!}$$

For which we can again find a closed form solution.

How about $$ \sum_{s=1}^\infty \frac{\lambda^{s-1}}{(s-1)!}s$$?

I can't find any trick here to factor out the $s$. Is there any other definition or feature I can use?

Answer 1

Hint

$$\sum_{s=1}^\infty \frac{\lambda^{s-1}}{(s-1)!}s=\frac{\mathrm d }{\mathrm d \lambda}\sum_{s=1}^\infty \frac{\lambda ^s}{(s-1)!}$$

Answer 2

$$\frac{\lambda^{s-1}}{(s-1)!}s=\frac{\lambda^{s-1}}{(s-1)!}(s-1)+\frac{\lambda^{s-1}}{(s-1)!}$$

Can you go on?

Answer 3

$$\sum_{s\geq 1}\frac{s}{(s-1)!}\lambda^{s-1} = \sum_{n\geq 0}\frac{n+1}{n!}\lambda^{n}=\frac{d}{d\lambda}\left(\lambda e^{\lambda}\right)=\color{red}{(\lambda+1)\,e^{\lambda}}.$$