Rudin defines a power series for the exponential function and says that $E(0)=1$. But when $z=0$, the first term in the series is $0^0$, which is undefined. Why is it equal to $1$?
3 Answers
$\begingroup$$0^0$ is not undefined; it is $1$.
However, $\lim a_n=0$ and $\lim b_n=0$ does not imply tat $\lim a_n^{b_n}=1$, but that is a totally different story - namely that $0^0$ is called an indeterminate form.
$\endgroup$ 1 $\begingroup$The first term of the development is formally
$$\frac{x^0}{0!}$$ which is indubitably defined to be $1$ for all $x\ne 0$. Then it is a natural convention (possibly left implicit) that this term is always $1$, making the function continuous while allowing a notational convenience.
Unless he stated this convention, the author would have been more careful to write
$$1+\sum_{k=1}^\infty\frac{z^n}{n!}$$ as @mathisfun said.
Note that this doesn't tell us anything about the value of $0^0$, if it has one.
$\endgroup$ $\begingroup$Defining $0^0 =1$ is pretty natural since $$\lim_{x\to 0+} x^x =1.$$
$\endgroup$ 1
