infinity - If $0.999cdots = 1$ Then Does $frac{1}{10^infty} = 0$?

Recently I stumbled across a, to me, rather strange idea. I was messing around with the proof of $0.999... = 1$, when I figured that what $0.999...$ means is that those are all nines. That way I came upon a weird idea. Say $a = 0.999...$, then $a = 1 - x$. Yet, how do we define what $x$ is? It should say $x = 0$, but my theory did not. If $0.999...$ is just an endless sequence of nines, then why can't we say $x$ just is an endless sequence of zeroes, ending with a 1, like $\frac{1}{10^\infty}$?
If we take the equation $$n = 0.9$$ for example, then, what would $1 - n$ be? Yes indeed, $0.1$. Following that theory, can't we say that $$0.999... = 1 - \frac{1}{10^\infty}$$Now, if $0.999... = 1$ this would be impossible. Go figure. $$1 = 1 - \frac{1}{10^\infty}$$ then $$\frac{1}{10^\infty} = 0$$ but that is impossible, because we cannot say $$10^\infty * 0 = 1$$When I came to this point, I really got stuck, because, in my head everything I did was right, however, it is impossible. Can somebody please explain to me what mistakes I may have made, and enlighten me about what else I did wrong?

Thanks in advance.
Sjoerd Dorrestijn.

EDT: I prefer $\frac{1}{10^\infty}$ to use as an indication of $0.000...01$, even though $10^\infty = \infty$ in some way, I just seem to find this more clear.

EDT2: Just to be clear, I read a proof that $0.999...9 = 1$ because it'd be $1 - 0.000...0 = 1 - 0 = 1$. What I tried to prove here is that it is not equal to $1$, because otherwise maths would collapse. My question was whether I am right or wrong. Since the original statement uses infinity (an infinite amount of nines) I think it is a must to use infinity as well. So, the question is if either the original statement is false, or if I made a mistake somewhere.

Answer

The (proven true) statement that $0.999\ldots=1$ means that $\lim_{n\to\infty}0.\underbrace{999\ldots9}_n=1$. That in turn is equivalent to $\lim_{n\to\infty}\left(1-0.\underbrace{999\ldots9}_n\right)=0$. Now note that $1-0.\underbrace{999\ldots9}_n=\frac{1}{10^n}$, so we have $\lim_{n\to\infty}\frac{1}{10^n}=0$.

However, although the above reasoning works, it is easier and more straightforward to conclude that $\lim_{n\to\infty}\frac{1}{10^n}=0$ directly from the definition of the limit, without ever considering $0.999\ldots$.

Finally, it is advisable to avoid the notation $\frac{1}{10^\infty}$ altogether, as it is not defined given the usual definition of its constituent symbols.