I was told that any $x^{\infty}$ is undefined. Does this hold true even when $x=1$?
If yes, why? $1$ to any power is always $1$? $\infty$ is not a number, but as the numbers get larger and larger, raising one to that power should still be one?
This really is all part of the question of what is $\lim_{x\to\infty}1^x$ and how to solve it. The limit is $1$, but it still leads to this question.
The literal expression "$1^\infty$" is undefined because infinity is not in the (relevant factor of the) domain of the power operation. Nothing in this expression is a limit or a sequence, so there is no sense in which something is getting larger and larger. This expression does not represent a process - it contains a "completed infinity".
The indeterminate form "$1^\infty$" appears when one is mentally approximating limits, commonly (but not exclusively) limits of the form $\lim_{x \rightarrow \infty} f(x)^{g(x)}$. The usual approach to such a form is to use the continuity of the exponential to write $$ \lim_{x \rightarrow \infty} \mathrm{e}^{g(x) \ln(f(x))} = \mathrm{e}^{\lim_{x \rightarrow \infty} g(x) \ln(f(x))} \text{.} $$
Now if $f$ is approaching $1$ in our limit, $\ln f$ is approaching zero, so the limit in that exponent is of the form "$\infty \cdot 0$", and we look for cancellation opportunities and other familiar manipulations to resolve the relative rates of $\ln f$ going to zero and $g$ going to infinity.
What does this do to the expression you wrote? We would write $\mathrm{e}^{\infty \cdot \ln 1} = \mathrm{e}^{\infty \cdot 0}$. But this does not resolve the issue: "$\infty \cdot 0$" is also undefined because completed infinities are not in the domain of multiplication. So this expression is also "$\mathrm{e}$ to an undefined power".