I was reading up on 0.999... = 1 and since my background is not in math, my mind started coming up with in my mind, intuitive, but misguided wrong problems. After some wolfram alpha i came to the following results:
1^infinite is either undefined or 1
0.999^infinite is 0
0.999...^infinite is undefined
Is there a more interesting answer for the last one than undefined?
http://m.wolframalpha.com/input/?i=1%5Einfinite&x=10&y=13 http://m.wolframalpha.com/input/?i=0.999%5Einfinite&x=9&y=8
It's really hard to answer this question without more context about what exactly you mean, but here is one way of thinking about it that might be interesting to you.
I'm going to assume that when you ask for "a more interesting answer", part of what you have in mind is thinking of the expression $0.999\dots$ as denoting not a single real number (which is the standard way of thinking about it) but rather as a placeholder designating a sequence of numbers $a_n$ converging to $1$, and when you say "infinite" you have in mind a sequence of numbers $b_n$ that grows without bound, and what you really want to know is: What is the limit of the sequence $a_n^{b_n}$?
The short answer is: Depending on how the sequences $a_n$ and $b_n$ are chosen, the result can turn out to be almost anything you want. But one particularly important example is to choose the sequences as follows: $$a_n = 1 - 1/10^n$$ $$b_n = 10^n$$
In that case the relevant sequence is $a_n^{b_n} = (1 - 1/n)^{10^n}$. The first few terms of this sequence are: $$0.9^{10}, 0.99^{100}, 0.999^{1000}, 0.9999^{10000} , \dots$$ and the limit of this sequence is $1/e$.
But other results are possible, too. If $x$ is any real number then the sequence $a_n = 1 + x/n$ also converges to $1$, and if you choose $b_n = n$ then the sequence $(1 + x/n)^n$ converges to $e^x$. By choosing $x$ appropriately, you can create a sequence that converges to any (positive) real number you like.