How to understand why $x^0 = 1$, where $x$ is any real number? 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?

RULE 1
$$\frac{x^m}{x^n} = x^{m-n}$$
thus if m=3 and n=3,
$$\frac{x^3}{x^3} = x^{3-3} = x^0$$
RULE 2
Anything devided by its self is 1, why?
$$\frac{10}{10} = 1,$$
because $\frac{y}{y} = y\left(\frac{1}{y}\right) = \left(\frac{y}{1}\right)\times \left(\frac{1}{y}\right)$ which can be displayed as
$$ \frac{y}{1} \times \frac{1}{y} = \frac{1}{1} = 1$$
here the $y$ values cancel out to become 1
COMBINING THE RULES
so if anything devided by it's self is 1 
and $x^0 = x^{m-n} = \frac{x^m}{x^n}$
then
$X^0$ must = 1, as its simply just a way of saying that value devided by itself =?
