Are there an $\aleph_0$ # of natural numbers with an
$\aleph_0$ # of digits?

 Every natural number has a finite number of digits.
 Adding to the above (which is correct), the set of infinite digit strings is uncountable.

ok I understand what you guys are saying but it still seems strange to me.
I feel like that is saying the natural numbers are not bounded but they have a finite number of digits. I mean you couldn't put a bound on the number of digits.

 Quote by cragar ok I understand what you guys are saying but it still seems strange to me. I feel like that is saying the natural numbers are not bounded but they have a finite number of digits. I mean you couldn't put a bound on the number of digits.
Each individual natural number has a finite number of digits.
The entire set is unbounded.

 I mean you couldn't put a bound on the number of digits.
You can't. This doesn't change the fact that every natural number has a finite number of digits.

 For any natural number you pick, I can pick one with more digits. For example, if you picked x I could pick 10x, or 100,000,000,000,000,000x. However all three of those numbers have a finite number of digits. As the natural numbers get larger and larger so do the number of digits. Say you have f(x) = # of digits x has for all natural numbers. Then it is certainly true that as x approaches infinity, so does f(x).
 To make the above a bit more rigorous, the number of digits in a natural number $n$ is given by $\lfloor \log_{10}(n) \rfloor$ and this obviously goes to infinity.
 I could see the problem with saying that there are natural numbers with an $\aleph_0$ of digits because then I would have 10 choices for each number in the slot and I would have $10^{\aleph_0}$ numbers which would be uncountable and a contradiction because the set of naturals is countable. Could I use this as a proof by contradiction to verify it?

 Quote by cragar Could I use this as a proof by contradiction to verify it?
No. The contradiction does not verify that every natural number has a base 10 representation with only finitely many digits.

Recognitions:
 Quote by cragar I could see the problem with saying that there are natural numbers with an $\aleph_0$ of digits because then I would have 10 choices for each number in the slot and I would have $10^{\aleph_0}$ numbers which would be uncountable and a contradiction because the set of naturals is countable. Could I use this as a proof by contradiction to verify it?
If I understood you correctly, you want to compose all strings of finite length

with terms in {0,1,..,9} . If you write those strings as

Ʃi=0Nai10i

and let N→∞ , then(a) problem is that your sum will diverge much of the time, so that

many of those strings are not natural numbers.

 ya thats what i am kinda saying