# How does $\exp(0)=1$ follow from the definition $\exp(z):= \sum_{n=0}^\infty \frac{1}{n!} z^n$

We introduced the Exponential function as follows: $$\exp: \begin{cases} \mathbb{C} & \longrightarrow \mathbb{C} \\z & \longmapsto \displaystyle \sum_{n=0}^{\infty} \frac{1}{n!}z^n \end{cases}$$ It might be a trivial thing to ask, but in my mind I always identify this function with $e^z$ or in the real domain with $e^x$. It is a very powerful definition because many interesting results in mathematics can be deduced from it in a straightforward manner, such as the Euler Identity and many trigonometric laws.

However I don't see how this definition is consistent with the fact that I remember from school about $e^0$ being equal to $1$. Maybe my computation is flawed but if I substitute $z=0$ I get.

$$\exp(0)=\sum_{n=0}^{\infty} \frac{1}{n!}0^n=\underbrace{0^0}_!+0^1+\frac{1}{2}0^2 + \dots$$ where $0^0$ is not defined, because $0^0=0^{-0} = \frac{0}{0}$

In this case $0^0$ is not defined if you look at it from the exponentiation (on integers) point of view, however $0^0:=1$ here. The reason for this is pragmatic: it facilitates notation.
It's not that the real number $0$ powered to itself equals $1$. No, that's not defined. In the context of series, $0^0=1$ just so it is easier to write stuff. You should translate $\sum \limits_{n=0}^{+\infty}\left(a_n0^n\right)$ to $a_0\cdot 1+\sum \limits_{n=1}^{+\infty}\left(a_n0^n\right)$
There are others instances in which $0^0=1$. For instance in cardinal arithmetic, given two finite cardinals $\alpha$ and $\beta$, $\alpha ^\beta$ denotes the number of functions from $\beta$ to $\alpha$. If $\alpha=\varnothing =\beta$, then $\varnothing ^\varnothing$ is the number of functions from the empty set on itself, and that's $1$, (it's the empty function). You don't even have to define $0^0$ separately here.
• Strange I haven't heard or read about this definition before, I believe all calculators I have stumbled across, including the wolf, identify $0^0$ is indeterminate wolframalpha.com/input/?i=0%5E0. So I believe $0^0:=1$ was made to 'save' this definition, fair to say? Commented Nov 16, 2013 at 14:53
• In the most general context, $\;0^0\;$ is undefined, but in some cases, as in the definition of the exponential function it is defined to be 1 for notational purposes and simplicity. Commented Nov 16, 2013 at 15:09