exponential null operator

Question

In $A$ banach algebra with unit, and $X\in A$. if i define

$e^X=\sum_{n=0}^{\infty} \frac{1}{n!}X^n$

why $e^0=Id$ , i am aassuming $O^0=Id $ with $0$ null operator

thanks

Answer 1

Since$$e^X=\operatorname{Id}+X+\frac{X^2}{2!}+\frac{X^3}{3!}+\cdots,$$then$$e^0=\operatorname{Id}+0+\frac{0^2}{2!}+\frac{0^3}{3!}+\cdots=\operatorname{Id}.$$