View 1:
Showing that $x^{x}$ approaches 1 as the positive value x gets arbitrarily close to zero does not prove that $0^{0} = 1$. The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that $0^{0}$ is undefined. $0^{0}$ does not have a value.
View 2:
For all $x>0$, we have
$$0^{x} = 0.$$
Hence,
$$\lim_{x \to 0^{+}} 0^{x} = 0$$
That is, as x gets arbitrarily close to $0$(but remains positive), $0^{x}$ stays at $0$.
On the other hand, for real numbers y such that $y \ne 0$, we have that
$$y^{0} = 1$$
Hence,
$$\lim_{y \to 0} y^{0} = 1$$
That is, as y gets arbitrarily close to $0$, $y^{0}$ stays at 1.
Therefore, we see that the function $f(x,y) = y^{x}$ has a discontinuity at the point $(x,y) = (0,0)$. In particular, when we approach $(0,0)$ along the line with $x=0$ we get
$$\lim_{y \to 0} f(0,y) = 1$$
but when we approach $(0,0)$ along the line segment with $y=0$ and $x>0$ we get:
$$\lim_{x \to 0^{+}} f(x,0) = 0$$
Therefore, the value of $$\lim_{(x,y) \to (0,0)} y^{x}$$ is going to depend on the direction that we take the limit. This means that there is no way to define $0^{0}$ that will make the function $y^{x}$ continuous at the point $(x,y) = (0,0)$.
View 3:
Zero raised to the zero power is one. No really, it’s true.
Let’s consider the problem of defining the function $f(x,y) = y^x$ for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:
$$y^x := 1 \times y \times y \cdots \times y$$
where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get
$$y^{x} = 1 \times y$$
However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving
$$y^{0} = 1$$
which holds for any y. Hence, when y is zero, we have
$$0^0 = 1$$
We've just proved that $0^0 = 1$ But this is only for one possible definition of $y^x$. What if we used another definition? For example, suppose that we decide to define $y^x$ as
$$y^x := \lim_{z \to x^{+}} y^{z}$$
In words, that means that the value of $y^x$ is whatever $y^z$ approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.Interestingly, using this definition, we would have
$$0^0 = \lim_{x \to 0^{+}} 0^{x} = \lim_{x \to 0^{+}} 0 = 0$$
Hence, we would find that $0^0 = 0$ rather than $0^0 = 1$. Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what $y^x$ means for all positive real numbers x and y, and it does preserve continuity of the function as we approach $x=0$ and $y=0$ along a certain line.
So which of these two definitions (if either of them) is right? What is $0^0$ really? Well, for $x>0$ and $y>0$ we know what we mean by $y^x$. But when $x=0$ and $y=0$, the formula doesn’t have an obvious meaning. The value of $y^x$ is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what $y^x$ means for positive values is not enough to conclude what it means for zero values.
But if this is the case, then how can mathematicians claim that $0^0=1$? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use $0^0=0$ or if we say that $0^0$ is undefined.
$$(a+b)^x = \sum_{k=0}^{\infty} \binom{x}{k} a^k b^{x-k}$$
Hence, if we do not use $0^0 = 1$ then the binomial theorem does not hold when $a=0$ because then $b^x$ does not equal $0^0 b^{x}$.
If mathematicians were to use $0^0 = 0$, or to say that $0^0$ is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to $k=0$. We gain elegance and simplicity by using $0^0 = 1$.
There are some further reasons why using $0^0 = 1$ is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more "natural" to mathematicians. The choice is not "right", it is merely nice.
