How can I prove that $\sum_{i=0}^{\infty} \frac{z^{i+1}}{(i+1)!} = e^z - 1?$

Question

How can I prove that $$ \sum_{i=0}^\infty \frac{z^{i+1}}{(i+1!)} = e^z - 1 $$

knowing that $$ \sum_{i=0}^\infty \frac{z^i}{i!} = e^z $$

Answer 1

Hint:

$$\sum_{i=0}^{\infty}\frac{z^i}{i!}=1+\sum_{i=0}^{\infty}\frac{z^{i+1}}{(i+1)!}$$ because $$\sum_{i=0}^{\infty}\frac{z^i}{i!}=1+\frac{z^1}{1!}+\frac{z^2}{2!}+\dots=1+\left (\frac{z^1}{1!}+\frac{z^2}{2!}+\dots\right )=1+\sum_{i=0}^{\infty}\frac{z^{i+1}}{(i+1)!}$$

Answer 2

Reindex: $$ \sum_{i=0}^\infty \frac{z^{i+1}}{(i+1)!}= \sum_{i=1}^\infty \frac{z^{i}}{i!}=\sum_{i=0}^\infty \frac{z^{i}}{i!}-1=e^z-1 $$