$1^\infty=2$

For any number $x$, we can express the difference between $x$ and $x$ raised to some power $n$ as:

$y=x-x^n$

To find which number $x$ most exceeds $x^n$, we can find the maximum point of the above function.

First the derivative:

$y\prime=1-nx^{(n-1)}$

And then the maximum (assuming that the power, $n$, is greater than 1):

$1-nx^{(n-1)}=0$

$nx^{(n-1)}=1$

$x^{(n-1)}=\frac{1}{n}$

$x=(\frac{1}{n})^{1/(n-1)}$

We can relabel this final equation as a function in $x$ and $y$ which will accept a power and yield the number which most exceeds itself raised to that power. In other words, a function which yields all the maximums of our original function ($y=x-x^n$).

$y=(\frac{1}{x})^{1/(x-1)}$

This function can be shown to converge to $1$:

$\lim_{x \to \infty}~(\frac{1}{x})^{1/(x-1)} = \lim_{x \to \infty}~(\frac{1}{x})^{0}=\lim_{x \to \infty}~1=\boxed{1}$

This is equivalent to saying that the number $n$ which most exceeds $n^{\infty}$ is $1$ ($1$ most exceeds $1^{\infty}$). Remember that (because it gets weirder!).

Now the function describing the difference between these "greatest" numbers and their respective powers is:

$y=(\frac{1}{x})^{1/(x-1)}-[(\frac{1}{x})^{1/(x-1)}]^x$

Simplifying:

$y=(\frac{1}{x})^{1/(x-1)}-[(\frac{1}{x})^{x/(x-1)}]$

$y=(\frac{1}{x})^{1/(x-1)}[1-(\frac{1}{x})^{(x-1)/(x-1)}]$

$y=(\frac{1}{x})^{1/(x-1)}[1-(\frac{1}{x})^{1}]$

$y=(\frac{1}{x})^{1/(x-1)}[1-(\frac{1}{x})]$

$y=(1-\frac{1}{x})(\frac{1}{x})^{1/(x-1)}$

This can also be shown to converge to $1$:

$\lim_{x \to \infty}~(1-\frac{1}{x})(\frac{1}{x})^{1/(x-1)} = \lim_{x \to \infty}~(1-0)(\frac{1}{x})^{0}=\lim_{x \to \infty}~1=\boxed{1}$

This is equivalent to saying that the difference between the number $n$ which most exceeds $n^{\infty}$ ($1$) and $n^{\infty}$ ($1^{\infty}$) is $1$.

This is understandably counter-intuitive, but taken in this context, it seems provable!

To finish, here is the grand consequence of this discussion:

$1^{\infty}-1=1$

Which means:

$\boxed{1^{\infty}=2}$

$\ddot\smile$