Look up indeterminate forms. It depends on how you interpret $1^{\infty}$
The indeterminate form $1^{\infty} = \lim_{n \rightarrow \infty} f(n)^{g(n)}$ where $\lim_{n \rightarrow \infty} f(n) = 1$ and $\lim_{n \rightarrow \infty} g(n) = \infty$
For instance, all of the following limits are of the form $1^{\infty}$, yet they all evaluate to different numbers
$$\lim_{n\rightarrow \infty} 1^n = 1$$
$$\lim_{n \rightarrow \infty} \left(1 + \frac{1}{n} \right)^n = e$$
$$\lim_{n \rightarrow \infty} \left(1 + \frac{1}{\log(n)} \right)^n = \infty$$
$$\lim_{n \rightarrow \infty} \left(1 - \frac{1}{\log(n)} \right)^n = 0$$
It is a question of which one tends faster whether $f(n) \rightarrow 1$ or $g(n) \rightarrow \infty$.
When you get a finite number which is non-zero and not one as the limit, there is in some sense a balance between the rate at which $f(n) \rightarrow 1$ and $g(n) \rightarrow \infty$.
When you get infinity or zero as the limit, $g(n) \rightarrow \infty$ faster than $f(n) \rightarrow 1$.
When you get $1$ as the limit, $f(n) \rightarrow 1$ faster than $g(n) \rightarrow \infty$.