Why is a number raised to the power zero always given to be equal to one?

Hint: An exponential function is mathematical function in form of \[{a^x}\] , where $x$ is a variable and $a$ is a constant which is called the base of the function and it should be greater that $0$ . To solve the given problem we use the property of exponential function.

Complete step-by-step solution:
First we find a number raise to some power.
After that we divide this number with same number.
e.g. if we get a number \[a\] with the power $x$ i.e.. ${a^x}$
now we divide this number with ${a^x}$ i.e., $\dfrac{{{a^x}}}{{{a^x}}}$
Now we know that $\dfrac{1}{{{a^x}}} = {a^{ - x}}$ , we use this and we get
${a^x} \times {a^{ - x}}$
We also know that ${a^x} \times {a^y} = {a^{x + y}}$ , use this in above operation and we get
$ = {a^{x - x}}$
$ = {a^0}$
Therefore $\dfrac{{{a^x}}}{{{a^x}}} = {a^0} = 1$
We how this for any number without zero
We know that $\dfrac{1}{0}$ is undefined.
Therefore this condition exists for all number except zero i.e., $a \in (0,\infty )$
As example we take ${3^x}$
Now we find the value of ${3^0}$
Therefore , we find the value of $\dfrac{{{3^x}}}{{{3^x}}}$
$\dfrac{{{3^x}}}{{{3^x}}} = {3^{x - x}}$
$ = {3^0}$
$ = 1$

Note: In the property of exponential function we use some conditions like \[\dfrac{1}{{{a^x}}} = {a^{ - x}}\] , ${a^x} \times {a^y} = {a^{x + y}}$ and ${a^x} \times {a^{ - y}} = {a^{x - y}}$ .
Use this properties and prove the required theory. For this problem we have to check the constant $a$ is greater then zero . If the constant $a$ equals zero then this theory cannot give the result. We cannot find the value of $\dfrac{1}{0}$ .