How to show that $\sum_{l=0}^{\infty} \dfrac{(l+1)^{n-1}}{l!}=\sum_{l=0}^{\infty} \dfrac{l^{n}}{l!}$ (proof of Dobiński's formula)?

Question

I am reading a proof of Dobiński's formula in Béla Bollabás book "The Art of Mathematics" (p. 144). There he uses

$$\frac{1}{e}\sum_{l=0}^{\infty} \dfrac{(l+1)^{n-1}}{l!}=\frac{1}{e}\sum_{l=0}^{\infty} \dfrac{l^{n}}{l!}$$

Is there an (easy) way to see why this should be true? I tried to shift variables but was not able to show the equality.

Answer 1

We have: $$\sum_{l=0}^{\infty} \dfrac{(l+1)^{n-1}}{l!} = \sum_{l=0}^{\infty} \dfrac{(l+1)(l+1)^{n-1}}{(l+1)l!} = \sum_{l=0}^{\infty} \dfrac{(l+1)^{n}}{(l+1)!} = \sum_{l=1}^{\infty} \dfrac{l^{n}}{l!} = \sum_{l=0}^{\infty} \dfrac{l^{n}}{l!}$$ since the term we add at the end is equal to $0$.