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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 $1$?

Picture of Rudin's definition

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    I think he puts $0^0 :=1$. – 2017-02-07
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    In calculus and analysis (where the exponent and base of an exponential expression are continuous), $0^0$ is undefined. In combinatorics, where both usually are constrained to integers, it makes sense to define $0^0 = 1$. For instance, if you have a code pad with no buttons on it, and you want to make a $0$-length code using that pad, how many possibilities are there? (Mis)using elementary combinatorics, we get the expression $0^0$, but counting directly, we get one answer: the empty code. This is somewhere in-between, but as long as the exponents are limited to integers, there's no problem. – 2017-02-07
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    It might be easier to think of the function as being defined by $$E(z) = 1 + \sum_{n=1}^\infty \frac{z^n}{n!}$$ so as to avoid the issue of giving $0^0$ a value. – 2017-02-07
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    Obviously $0^0$ isn't defined using the logarithm, but most definitions that do not use analysis to define powers imply that$0^0 =1$. I'll cite the ones that I have in mind : the definition by induction: you want to define $n^m$ inductively on $m$, uniformly on $n$ -> you have to start with $n^0 =1$; the definition through the number of mappings : the most elementary (in that it doresn't require many objects) definition of $n^m$ is the cardinality of the set of mappings from a set of cardinality $m$ to a set of cardinality $n$. Well when you do that with $n=m=0$, you get $1$. – 2017-02-07

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$0^0$ 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.

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    Isn't this a little contradictory then ? What does $0^0$ denote ? – 2017-02-07
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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.

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Defining $0^0 =1$ is pretty natural since $$\lim_{x\to 0+} x^x =1.$$

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    In the case on hand, it is more relevant to use $\lim_{x\to0}x^0$. – 2017-02-07