There are many different interpretations of division in elementary mathematics:

The expression $\frac{a}{x}$, or $a\div x$, can be interpreted as:

- The number of times that $x$ goes into $a$. For example, $\frac{10}{5}=2$ because $5$ goes into $10$ twice. The symbolic way of putting this is $\frac{a}{x}=q \equiv xq=a$ (here the symbol "$\equiv$" denotes equivalence). With this interpretation, $\frac{x}{x}=1$ because $x$ goes into $x$ once, or because $x1=x$.
- The number that goes into $a$ $x$ times. For example, $\frac{10}{5}=2$ because $2$ goes into $10$ $5$ times. The symbolic way of putting this is $\frac{a}{x}=q \equiv qx=a$. With this interpretation, $\frac{x}{x}=1$ because $1$ goes into $x$ $x$ times, or because $1x=x$. This interpretation is the same as interpretation 1 because $qx=xq$, the commutativity of multiplication.
- The ratio of $a$ and $x$, i.e. $a$ per $x$. This can be thought of best in terms of rates. Think of two objects, $A$ and $X$, traveling along the number line at constant velocity. $A$ travels from $0$ to $a$ in the same time interval in which $X$ travels from $0$ to $x$. Then $\frac{a}{x}$ is the distance that $A$ travels for each unit distance that $X$ travels, or the distance that $A$ travels in a certain time interval divided by the distance that $X$ travels in that same time interval. For example, $\frac{10}{5}=2$ because in this case $A$ travels a distance of $2$ every time $X$ travels a unit distance, and in any time interval, $A$ will travel twice as far as $X$. With this interpretation, $\frac{x}{x}=1$ because $X$ will travel a distance of $1$ for each unit distance that an object of identical trajectory travels.
- $a$ $x$ths. For example, $\frac{10}{5}$ is $10$ fifths, which is $2$. The symbolic way of putting this is $\frac{a}{x}=a\times\frac{1}{x}$. With this interpretation, $\frac{x}{x}=1$ because $x$ $x$ths of something is $1$.

The above list is not exhaustive; there are other ways of interpreting division.

However, you don't need division to show that $x^{0}=1$. Here is an alternative way.

Let $S$ be a multiset of numbers (a multiset is a set in which elements can appear more than once), for example $\left\{2,3,3,6\right\}$. Let $\Pi(X)$ denote the product of all the numbers in the set $X$, and let $\Sigma(X)$ denote the sum of all the numbers in the set $X$. When adding and multiplying, the order and grouping of the terms or factors has no effect on the result (this follows from an inductive argument on the number of terms/factors that uses the associative and commutative properties). Thus if $S$ is separated into two disjoint sets $S_{1}$ and $S_{2}$ (for the above example, these could be $S_{1}=\left\{2,3\right\}$ and $S_{2}=\left\{3,6\right\}$), it is the case that $\Pi(S)=\Pi(S_{1})\times\Pi(S_{2})$ and $\Sigma(S)=\Sigma(S_{1})+\Sigma(S_{2})$. The result about the order and grouping of terms/factors only entails this if $S_{1}$ and $S_{2}$ are nonempty, but it would make sense to define $\Pi(X)$ and $\Sigma(X)$ such that it works for all sets $S_{1}$ and $S_{2}$, even empty ones. Let $S_{1}$ be the empty set, denoted $\varnothing$. Then $S_{2}=S$. We want to have $\Pi(S)=\Pi(\varnothing)\times\Pi(S)$ and $\Sigma(S)=\Sigma(\varnothing)+\Sigma(S)$, so we must define $\Pi(\varnothing)=1$ and $\Sigma(\varnothing)=0$.

What we have shown is that the sum of all the elements in the empty set is $0$ and the product of all the elements in the empty set is $1$. When we write $x^{n}$, we are taking the product of all the elements in the multiset that has $x$ in it $n$ times and nothing else. So $x^{0}$ is the product of all the elements in the empty set, or $\Pi(\varnothing)$. Therefore $x^{0}=1$. This argument can also be used to show that $0!=1$, as this is also $\Pi(\varnothing)$.