How can $x^0=1$ be proved with a real life experiment? I saw that $x^0=1$ proof is $\displaystyle x^{a-a} = \frac{x^a}{x^a} = 1.$
However I don't find this a convincing proof.
How can someone prove this without using the fraction above? or, Is that fraction the only proof of $x^0=1$?
 A: With compound interests, your capital increases by $10\%$ every year.
$$1 \to 1.1\\
2 \to 1.21\\
3 \to 1.331\\
4 \to 1.4641\\
\cdots\\
N\to1.1^N
\cdots$$
Obviously we get a consistent definition if
$$0\to1$$

A: You can only prove mathematical statements with arguments that start from mathematical axioms. "Real life" does not enter the picture.
That said, the assertion that for $x>0$
$$
x^0 = 1
$$
is not a theorem, it's a definition. Mathematicians could have decided that
$$
x^0 = 17, \text{ or } 0, \text{ or } \ldots
$$
but chose $1$ because it turns out to be the most useful.
You are entitled to an explanation for that choice.
Remember that for ordinary positive integers,
$$
x^m  x^n = x^{m+n}
$$
(true because $x^n$ is defined as the result of multiplying $x$ by itself $n$ times).
That rule for adding exponents is so useful that mathematicians decided to preserve it when extending the meaning of raising $x$ to a power. Then
$$
x^0  x^n= x^{0+n} = x^n . 
$$
The only way that can happen is if you define $x^0$ to be $1$.
The same desire to preserve the rule for adding exponents is why
$$
x^{-1} = \frac{1}{x} \text{ and } x^{1/2} = \sqrt{x}.
$$

For another good reason to define the product of no numbers as $1$ see Empty set and empty sum
A: If your notion of exponentiation is ultimately based on repeated multiplication (there are other ways to approach powers, where the answer may differ), then $x^0=1$ cannot be proven. It is defined to be such. And your fraction is one of many, many reasons that $1$ is the most reasonable value to give $x^0$, by a wide margin.
A: Start with the following equation
$$x=a, $$
in which $a>0$ and apply the $n^\text{th}$-root.
$$x^{\frac 1n}=a^ {\frac 1n}$$
If you take the limit $n \to \infty$ you will obtain
$$x^0=1,$$
because the infinith root of a positive number is $1$.
