For $x\in\mathbb R$ we define $$\exp(x) := \sum_{n=0}^\infty \frac{x^n}{n!}. $$ This is the standard definition of the exponential function, e.g. given by Rudin in the introduction to Real and Complex Analysis. It is well-known that this series converges absolutely for all real $x$, so this definition is perfectly acceptable.
I have a rather silly question though. How do we get $$\exp(0)=1 $$ from this definition? It's clear if we use a different definition of $\exp$: $$\lim_{n\to\infty}\left(1+\frac0n\right)^n = \lim_{n\to\infty}1^n=1. $$ But how do we reconcile the expression of the form $0^0$ in $$\exp(0) = \sum_{n=0}^\infty \frac{0^n}{n!} $$
My thoughts: if $x\ne0$, then $x^0=1$, so $\lim_{x\to0}x^0=1$. Hence for any nonnegative integer $N$, $$\lim_{x\to0}\sum_{n=0}^N\frac{x^n}{n!} = 1,$$ and so $$\lim_{N\to\infty}\lim_{x\to0}\sum_{n=0}^N\frac{x^n}{n!}=1. $$ It remains to show that the above is equal to $$\exp(0) = \lim_{x\to0}\exp(x)=\lim_{x\to0}\lim_{N\to\infty}\sum_{n=0}^N\frac{x^n}{n!}. $$ Is the absolute convergence of $\exp(x)$ for $x\ne0$ and the uniform convergence of $\exp$ on any bounded subset of $\mathbb R\setminus\{0\}$ enough to justify the above interchange in limits? If so, post it as an answer, and I will accept it and put this admittedly pedantic question to rest.
The standard convention is that when you write $x^n$, then when $n = 0$, this is the constant function $1$, regardless of the value of $x$. This is what people always mean by $x^0$ in a power series $\sum_{n \ge 0} a_n x^n$. This is the unique convention, for example, that makes $x^0$ continuous from the right as a function of $x \in \mathbb{R}_{\ge 0}$.