When typing a calculation to Wolfram Alpha it always takes $0^0$ as undefined, but for some calculations I need to preform, I specifically need it to be equal $1$.
Is there a way to make it understand $0^0$ as $1$, or is there at least a workaround for this? (cases of $0^n$ where $n>0$ should still be $0$)
To me $0^0$ is very nicely defined as the cardinality of the set of maps from the empty set to the empty set, hence equals $1$. However, in the context of limits one must be aware that the binary operation of exponentiation is not continuous at $(0,0)$ (and not even defined in an open neighbourhood of $(0,0)$), which implies that $a_n\to 0$, $b_n\to 0$ gives us no idea what might happen to $a_n^{b_n}$ as $n\to\infty$. Therefore $0^0$ is called an indeterminate form, just like $\frac 00$; however the latter is not only an indeterminate form but also undefined. As this is the most-encountered specimen of indeterminate form, it is not surprising that confusing "indeterminate" and "undefined" is wide-spread.
Note that "indeterminate form" is really about the unevaluated expression $0^0$. Normally, if two things are equal, they have the same properties, so if $0^0$ is indeterminate and $0^0=1$, we conclude that $1$ is indeterminate - which it is of course not. The important detail is that we are not talking about the value of $0^0$ as being indeterminate, but rather the "syntactic" expression (and that's why those things are called indeterminate forms, not indeterminate values)
To clarify, we say that the ("syntactic") form $a\circ b$ is an indeterminate form if $a_n\to a$ and $b_n\to b$ does not allow conclusions about the existence or value of $\lim_{n\to\infty}a_n\circ b_n$.
Thus writing something like $$ \lim_{n\to\infty}a_n^{b_n}=0^0=1$$ is probably wrong because $a_n\to 0$, $b_n\to 0$ does not warrant the first equality. In contrast, writing $$ \lim_{n\to\infty}\frac{a_n}{b_n}=\frac00=?$$ is definitely wrong because you cannot equate something with an undefined expression.
Similarly to $0^0=1$, it is customary in some branches of math to define $0\cdot \infty=0$. But being defined does not mend the fact that $0\cdot \infty$ is an indeterminate form in the above sense.
The following is taken from this answer.
In common usage, $0^0$ is often encountered in set theory as the number of maps from the empty set to the empty set, or as $x^0$ in combinatorics and polynomials. In all of these cases, $0^0=1$ is the proper definition, since there is $1$ map from the empty set to the empty set, and because $$ \lim_{x\to0}x^0=1 $$ Certainly, there are limits of the form $0^0$ which do not equal $1$, for example, $$ \lim_{x\to0}|2x|^{1/\log|x|}=e $$ But they do not occur as often as those mentioned above. Furthermore, since there is a problem raising negative numbers to non-integer powers, even defining $x^y$ in a neighborhood of $(0,0)$ is difficult. This is why we usually consider $x^0$, where the exponent is a fixed integer, when talking about $0^0$.
This has been a subject of debate, and there is really no one resolution.
One reason for this is that there is no one definition of exponentiation itself. There are several definitions corresponding to different domains of the exponent. We can prove (or at least, render) these definitions equivalent where the domains overlap for strictly positive bases, and we might be tempted to suppose that the equivalence can be carried over to other bases by "analytic continuation". But it can't because exponentiation does not in general give an analytic function.
The result is that some definitions lend themselves to rendering $0^0$ as $1$ and others do not. Ultimately the validity of making that assumption depends on the context in which the exponent arises, for that context detetmines which definition of exponentiation is relevant.
The following discusses this problem much more extensively.
As they probably introduce exponentiation on the natural numbers in most schools, for any natural number $n$, we have:
$n^2 = n\times n$
$n^3 = n\times n \times n$
$n^4 = n\times n \times n\times n$
$\vdots$
$n^{m+1} = n^m\times n\space\space\space$ for $m\ge 2$
$\vdots$
Now, if we drop the restriction that $m\ge 2$, it turns out that this scheme can be used to determine $n^m$ for all natural numbers $n$ and $m$, except for $n=m=0$. In fact, any value will work for $0^0$ in this scheme.
You can pick any convenient value for $0^0$. The values $0$ or $1$ result in some nice outcomes. Or you can leave it undefined as they probably taught you in school.
