Defining $0^0=1$ I've read in several places that defining $0^0=1$ is convenient in several (primarily discrete) settings. One argument on Wikipedia in favor of this definition was the need of a special case for the product rule $\frac{d}{dx} x^n = nx^{n-1}$ for $n=1$ at $x=0$.
Regardless of which definition, if any, is proper, is this not an argument to the contrary? Or I'm missing something? $nx^{n-1}$ for $n=1$ at $x=0$ is $1*0^0 = 1*1 = 1$ contradicts $\frac{d}{dx} 0^1 = \frac{d}{dx} 0 = 0$.
I feel as though there's something just incredibly obvious I'm missing here. How can this be an argument in favor of defining $0^0=1$?
 A: There is actually a fairly good motivation for this definition, but it requires set theory, and a bit of torturing of definitions.
In set theory, given two sets $X,Y$, the set $X^Y$ is defined to be the set of all maps from $Y$ to $X$.  Let $|X|$ be the cardinality ("size") of $X$ as usual.  Then $|X^Y|$ = $|X|^{|Y|}$ if $|X|$ and $|Y|$ are positive integers.  But if $|X|=|Y|=0$, so $X$ and $Y$ are both empty sets $\emptyset$, then there is one map; from the empty $Y$ to the empty $X$!  In other words, $|\emptyset^\emptyset|=|\emptyset|^{|\emptyset|}=0^0=1$.
Now in order for this motivation to work fully, to have to declare that only the empty set can map to the empty set.  Then $|X^\emptyset|=1$ for all $X$, because there is only the one map from $\emptyset$ to $\emptyset\in X$.  Further, $|\emptyset^Y|=0$ if $Y\ne\emptyset$, because then there is no empty set to map from.
Like I said, a bit of torture, but it works!
A: Review the summary discussion on this topic here: http://mathworld.wolfram.com/Zero.html
Things tend to get messy at times!
Regards -A
