TITLE: How to understand why $x^0 = 1$, where $x$ is any real number?
QUESTION [23 upvotes]: Alright, so the idea of an exponent, $x$, is that you are multiplying its base by itself $x$ number of times. With base $5$ and $x=3$, we have that $5^3$ = $5 \cdot 5 \cdot 5$
I understand that the logarithm with base $a$ of $x = c$,
tells us that
$$a^c = x$$
and for $c =$ positive; values for $x$ are greater than $1$, and for $c =$ negative; values for $x$ are less than $1$,  and for $c = 0$, values for $x$ are...$1$.
So in short, I understand how, by means of observation of the graph of $f(x) = \log x$, we can see that $f(1) = 0$, BUT, I see no other way to understand why $x^0 = 1$, apart from the graph and everything around that very point.
I honestly cannot get my head around the idea, "$5$ times itself $0$ times... is one".
Is it that there is no fundamental answer for this but that we simply know by the graph? Or can I truly understand $x^0 = 1$ on its own?

REPLY [4 votes]: I see that this question already has seventeen answers, but I'm going to jump on the bandwagon because as a set theorist I feel very underrepresented by the current answers.
For $m,n \in \mathbb{N} = \{0,1,2,\dots\}$ we can define $m^n$ to be the number of functions from a set of size $n$ to a set of size $m$. For instance,
$$3^5 = \text{the number of functions}\ \{1,2,3,4,5\} \to \{1,2,3\}$$
There's only one set of size $0$, namely the empty set $\varnothing$, and so for any $m \in \mathbb{N}$
$$m^0 = \text{the number of functions}\ \varnothing \to \{1,2,\dots,m\}$$
But for any value of $m$ there is only one such function, namely the empty function. (This is a function: its domain is empty so it satisfies the definition trivially.) Therefore $m^0 = 1$ for all $m \in \mathbb{N}$.
This definition of exponentiation extends naturally to $\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \cdots$ and also to infinite cardinal numbers.
More interestingly, according to this definition, we have $0^0=1$ (since there is exactly one function $\varnothing \to \varnothing$) without any fear of lack of definition. (It does mean that the function $x \mapsto 0^x$ is discontinuous at $0$, but who cares? It means that the power series definition of $\exp x$ is well-defined at $0$, which I think trumps this caveat!)