Why is everything (except 0) to the power of 0 always 1?

Question

Why does $x^0=1$? Also, why does this rule not apply when $x=0$? I understand that math can be difficult to explain, but this subject never seems to be thoroughly explained in simple terms.

Answer 1

$$x\ne 0$$

$$x^0 = x^{a-a}$$

$$x^0 = \frac {x^a}{x^a}$$

$$\frac {x^a}{x^a}=1$$

$$x^0=1$$

Answer 2

Say we have a value $x$ and we want to raise it to a power $n$. This means that we multiply $x$ by itself $n$ times. $$x^n = \underbrace{x\cdot x\cdot x\cdot\ldots\cdot x}_{n\text{ times.}}\tag1$$ It is confusing to most people when we say that $x^0 =1$, because how can we multiply $x$ by itself $0$ times? Here, we have to look at one of the Power Rules.

The first power rule is as follows: $$x^a\cdot x^b = x^{a + b}.$$ This is provable from $(1)$. Since $x^a$ and $x^b$ are all products of $x$, then when we multiply them together, the number of times $x$ is being multiplied by itself in total is of course $a + b$ times. Therefore, the product of $x^a$ and $x^b$ is always $x^{a+b}$.

So this means that since $n = n + 0$, we get that $x^n = x^n\cdot x^0$. Therefore, $x^0$ must be equal to $1$. $$x^0 = 1.$$ But what happens when $x = 0$? Well here, we have to look at the second power rule. $$\frac{x^a}{x^b} = x^{a-b}.$$ This is also provable using the same method we used to prove the first power rule. But this means that when $x = 0$, we introduce division by $0$ which cannot exist (for obvious reasons). Since every number $x$ to the power of $1$ is equal to itself (this is also a power rule) then we can write $0^1 = 0$. Now if we want to solve for $0^0$, well according to $(1)$, we have that $0^0 = 1$. But, $$0^0 = 0^{1-1} = \frac{0^1}{0^1} = \frac{0}{0}.$$ This is where division by $0$ is introduced, and is why $0^0\neq 1$. But since $0 - 0 = 0$ then, $$\frac{0}{0} = \frac{0^0}{0^0}.$$ Therefore, because we cannot have division by $0$, we get that $0^0\neq 0$. Now looking back at the first power rule, since $0\times 0\times 0\times\ldots\times 0 = 0$ then we get that $0^n = 0$, for all $n > 0$. Of course, if $n < 0$ then this would introduce division by $0$ as well. Furthermore, this means that $x^0 = 1$, but we need $x\neq 0$.

To conclude, we get that $0^0$ is just a special case of a power of $0$, and it is thus undefined. For all $x\neq 0$, we can define $x^0$ as simply being equal to $1$. And for all $n > 0$, we can define $0^n$ as simply equal to $0$.