Why Is $1^{\infty}$ an indeteminate form?

Question

For me, multiplication is a binary operation so it can be applied only on a finite sequence of numbers. but $1^{\infty}$ requires that we apply multiplication infinitly which is not defined as multiplication is a binary operation.

Is that a good reason? If not, what is the reason?

If my reason is ok, So similarly, $5^{\infty}$ is indeterminate ?

ِAdded:

I noticed that all answers are in context of "limits". Algebraically, multiplication is a binary operation, So it ONLY can be used to define multiplication of finite sequence of numbers as not a infinite sequence. So algebraically, what does $1^{\infty}$ even mean?

Answer 1

The reason on why this is indeterminate is because of how it behaves when we go to limits. For example, if you look at $1^n$ as $n\rightarrow\infty$ we would get that $1^\infty=1$, while by looking at $(1+\frac{1}{n})^n$ we get that $1^\infty=e$. We don't get such thing with $5^\infty$ because, no matter what, it always diverges to infinity.

Answer 2

Because $1=a^0$, and $0\cdot\infty$ is $($also$)$ undetermined.

Because all convergent infinite products are of the form $1^\infty$, since their general term tends to $1$, and the number of terms is infinite, but they don't all converge to the same value. Furthermore, there are also divergent infinite products whose general term also tends to $1$.

Etc.