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

$x^{n+1}=x\cdot x^n$ right?

so

$x^1=x \cdot x^0$ but $x=x^1$ so for that to hold true, $x^0$ must be $1$.

Similarily,

$\large x^{-n} = \frac{1}{x^n}$.

So $\large x^n \cdot x^{-n} = x^n \frac{1}{x^n} = 1$.

But $\large x^n \cdot x^{-n} = x^{n+(-n)} = x^0$, so once more, $x^0=1$.

There are really many reasons for that to hold, and all of them are just a consequence of some agreements we've made previously.

Don't think of $x^n$ as $\overbrace{x\cdot x\cdots x}^{n\text{ copies}}$. Instead think of $x^n$ as representing the result of starting with $1$ and applying "${}\cdot x$" $n$ times: $1\overbrace{{}\cdot x\cdot x\cdot{}\cdots{}\cdot x}^{n\text{ copies}}$.

This also makes negative exponents a little nicer for beginners. $x^{-n}$ is just applying the opposite of ${}\cdot x$, which is ${}\div x$, $n$ times. $x^{-n}=1\overbrace{{}\div x\div x\div\cdots\,\div x}^{n\text{ copies}}$.

There are many answers about why $x^0 = 1$ for general $x$, so I'd like to address a different issue here, the way you think about exponentiation, which seems to be troubling you.

The definition you use for exponentiation holds true for integers, and rationals if you define what $x^{1/n}$ means, but what about the irrationals? How can you times something by itself $\sqrt{2}$ times? The answer is - you can't! You need to come up with some sensible way to define exponentiation rigorously.

We define $$e^x = \displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$$ which is absolutely convergent for any complex $x$, and only depends on us defining what $x^n$ means for the natural numbers and $0$. Considering the function restricted to the reals we can use this definition to show that the function is continuous, differentiable, has an inverse $\ln{x}$ and satisfies all the properties that we are used to with exponentiation. To then define, $a^x$ for $a \not= e$ we say $a^x = e^{x\ln(a)}$. This obeys all the properties that we want it to for rational $x$ and has the added bonus of being well defined on the irrationals too.

In fact, the definition of $e^x$ can even be seen as a motivation to define $x^0$ as $1$. $x^n$ for any natural $n$ is defined intuitively, but choosing any value for $x^0$ which is not $1$ will mean that this definition of $e^x$ will not longer obey the rules we have come to know and love e.g. $e^{a+b} = e^{a}e^{b}$. When you look at it like this, you really have no choice!

I hope you find this useful!