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$1^\infty$ = undefined
$2^\infty = \infty$
$0^\infty = 0$
Why is $1^\infty$ undefined? People were trying to explain to me that infinity isnt part of the Real numbers, yet, $2^\infty$ and $0^\infty$ somehow ARE defined?
In my opinion $1^\infty = 1$. I mean isn't it not easy to prove since $1\times 1\times 1 \times \cdots=1$ no matter how many times you do it?
It's best not to think of $1^\infty$ as a value, but rather as a shorthand - the same can generally be said anywhere that you see $\infty$ in this sort of analytical context. For instance, $0\cdot\infty$ (usually) isn't the actual product of these two entities, but rather a shorthand for an expression of the form $\lim\limits_{n\to\infty}(a_n\cdot b_n)$, where $\lim\limits_{n\to\infty}a_n=0$ and $\lim\limits_{n\to\infty}b_n\to\infty$ (that is, $b_n$ grows unboundedly). Similarly, $\dfrac00$ usually refers to an expression of the form $\lim\limits_{n\to\infty}\left(\dfrac{c_n}{d_n}\right)$ with $\lim\limits_{n\to\infty}c_n=0$ and $\lim\limits_{n\to\infty}d_n=0$; etc.
Part of the reason that the shorthand works is that it preserves the maning of any well-defined expression (in fact, that can be used as a definition of well-defined); we can equally well say that $1\cdot 2=2$ 'in shorthand' because if $\lim\limits_{n\to\infty}a_n=1$ and $\lim\limits_{n\to\infty}b_n=2$ then $\lim\limits_{n\to\infty}(a_n\cdot b_n)=2$ also. When people say that $1^\infty$ isn't well-defined, what they're really saying is that this shorthand falls down and doesn't give us a unique value.
These rules encode information about limits of sequences. If you are trying to evaluate $\lim_{n \to \infty}(a_n)^{b_n}$ and are told $a_n \to 0, b_n \to \infty$ you can be sure the limit is zero. Similarly, if you are told $a_n \to 2, b_n \to \infty$, you can be sure the limit is $\infty$. If you are told $a_n \to 1, b_n \to \infty$, you don't have enough information. In that way it is similar to $\lim_{n \to \infty}\frac{a_n}{b_n}$ where $a_n,b_n \to 0$. You need more information to evaluate the limit. It could be $0$, it could be some finite number, it could be $\infty$
$1*x=x$ for any $x \in \mathbb{R}$, but that is true only if the "$1$" you're talking about is exact, not in a limit sense.
Consider this famous limit $$\lim_{n \to \infty} (1+\frac{1}{n})^n$$
This is $1^{\infty}$, but it's limit is a very important mathematical constant denoted by $e$.
You can convince yourself that $e \neq 1$ by calculating that limit using a calculator. Try to plug in positive integers for $n$ in $\displaystyle A_n= (1+\frac{1}{n})^n$ and you'll see that the value of $A_n$ is getting bigger and gets closer and closer to $2.718\cdots$
Because $+\infty$ is an element of the extended real numbers. That number system is extremely useful in calculus and analysis, which is why people explain things using them.
Continuity is a key idea in these fields, and we often prefer to continuously extend functions whenever possible. e.g. we often like to continuously extend the function $f(x) = x/x$ to have the value $f(0) = 1$. And when we use the extended real numbers, we like to continuously extend all sorts of functions to have values at $+\infty$ and $-\infty$ when appropriate.
For any $x > 1$, you have $x^{+\infty} = +\infty$. This is right because, for example,
$$ x^{+\infty} = \lim_{\substack{y \to x \\ n \to +\infty}} y^n $$
And for $0 < x < 1$, you have $x^{+\infty} = 0$ for the same reason. (this post will not comment on defining $0^x$)
But because $1$ lies on the boundary between where the value of $x^{+\infty}$ should be $0$ and where it should be $+\infty$, you can't continuously extend the function $x^n$ to $x = 1$ and $n = +\infty$. Thus, we leave it undefined.
As an aside, if you were thinking specifically of $f(x) = 1^x$ rather than $g(x,y) = x^y$, then it would make sense to define $f(+\infty) = 1$, even though we need to leave $g(1, +\infty)$ undefined.
I would have thought if anything $1^\infty$ was defined as anything to the power of $1$ is $1$? The same for $0^\infty$ as $0$ to the power of anything is still $0$. However $2^\infty$ would be undefined as the real numbers are an open set so it just gets infinitely larger as your power tends towards $\infty$?
You can think in the limits of $0^n$, $1^n$ when $n$ tends to infinity. And with $2^n$ you can think in a unbounded positive sequence.
Consider $u_\epsilon \to a>1$ and $v_\epsilon\to b<1$, and $A_\epsilon\to\infty$.
Then $u_\epsilon^{A_\epsilon} = \exp [A_\epsilon \log u_\epsilon] $;
$\log u_\epsilon\to\log a>0 $ then $$ A_\epsilon \log u_\epsilon\to\infty \\ u_\epsilon^{A_\epsilon}\to\infty $$
If you replace $u_\epsilon$ with $v_\epsilon$ you get
$$ A_\epsilon \log v_\epsilon\to -\infty \\ v_\epsilon^{A_\epsilon}\to 0 $$
Since it seems like no one else has emphasized this:
$1^{\infty}$ is not well defined because $\infty$ is not a number.
When we do $2^3$ we are multiplying $2$ by itself $3$ times and we get the real number $8$. We extend the definition of $a^b$ so that it works for any positive real number $a$ and any real number $b$. And $\infty$ is not a real number. So $a^\infty$ is not well defined no matter what $a$ is.
It is the same with $\infty + 2$. Again this is not defined because $\infty$ is not a number.
Likewise $2^{\infty}$ is not well defined.
Here is the catch though: We can make sense of things like $\infty + 2$ by talking about the extended real number line. We can also make sense of $\infty$ when we talk about limits.
So for example $$ \lim_{x\to 0^+} \frac{1}{x} = \infty. $$ Things like this have precise definitions. And it is tempting to write the above as $$ \frac{1}{0}. $$ But this is not defined. We are only allowed to divide by non-negative real numbers.
That said, you can find various online sites that will say that $\frac{1}{\infty} = 0$. What they mean is that $$ \lim_{x\to \infty} \frac{1}{x} = 0. $$ This means that you can't necessarily trust these online sites when you are trying to carefully write down a mathematical argument.
Now to try to get at what is going on with saying that $1^{\infty}$ it is also worth pointing (as is done in other answers) that we can interpret $1^{\infty}$ differently in the context of limits.
For example consider: $$ \lim_{x\to 1^+} x^{\frac{1}{1-x}}. $$ Here the base approaches $1$ and the exponent approaches $\infty$. If we want to actually calculate this limit, we have to work hard than just appealing to some general definition of $1^{\infty}$. We can solve the problem using logarithms: $$ y = x^{\frac{1}{1-x}}\\ \ln(y) = \ln(x^{\frac{1}{1-x}}) = \frac{\ln(x)}{1-x} $$ and then find (using L'Hopital's rule) $$ \lim_{x\to 1^+} \ln(y) = \lim_{x\to 1^+}\frac{\frac{1}{x}}{-1} = -1. $$ So $$ \lim_{x\to 1^{+}} y = e^{\ln(y)} = e^{-1}. $$ Likewise it is possible to find similar examples where the limit is any given real number. You can even make the limit infinity.
