A "number" with an infinite number of digits is a natural number? The set of natural numbers is infinite and countable. Ok. But think of an object with infinite digits (141258173412873....). Is it a natural number?

Edit: What i found confusing was the fact that, since $\mathbb{N}$ is an infinite set, an object with infinite digits should be also a number and should belong to $\mathbb{N}$. I know this is a naive view. But now things are clearer to me, thanks to your answers! If i had to explain to a person not (too) educated in mathematic what $\mathbb{N}$ (the set of natural numbers) is, i would start with this:
consider the following algorithm (procedure) to construct $\mathbb{N}$={1,2,3,4....}:


*

*num = 1

*$\mathbb{N}$ is the empty starting set of numbers

*put num in $\mathbb{N}$

*num = num + 1

*repeat from 3


Now, does $\mathbb{N}$ has objects with an infinite number of digits? No. The procedure goes on forever,  but everytime we add a number to $\mathbb{N}$ (step 3), the number we are adding has a finite number of digits.
This view is only slightly different from other answers given to my original question, but i think it is simple enough to explain why a procedure that goes forever and build objects with an increasing number of digits does not produce a set with objects with an infinite number of digits.
 A: Let us consider the set of all "objects with infinite digits" and call it $S$.  You have said that $\mathbf{N}$ is countable.  Is $S$?  If you consider Cantor's diagonal argument, you will quickly see that $S$ is an uncountable set.  Thus, at the very least, there are infinitely more elements of $S$ than there are natural numbers.
A: The principle of mathematical induction says that if $S$ is a subset of the natural numbers such that:


*

*$1\in S$; and

*If $n\in S$, then $n+1\in S$;


then $S=\mathbb{N}$; that is, $S$ is the set of all natural numbers.
Let $S=\{k\in\mathbb{N}\mid k\text{ has only finitely  many digits when written in base 10 with no leading 0s}\}$.
Clearly, $1\in S$. If $n\in S$, then $n$ can be written with a finite number of digits, say $k$. Then $n+1$ can be written with either $k$ digits as well, or, in the worse case scenario (when $n = \underbrace{9\cdots99}_{k\text{ digits}}$) with $k+1$ digits; either way, if $n\in S$ then $n+1\in S$.
By induction, we conclude that $S=\mathbb{N}$. That is, every natural number has only finitely many digits when written in base 10 with no leading 0s.
So an infinite string of digits (omitting the "silly" possibility of infinitely many leading 0s), whatever it may be, is not a natural number.
A: Let me first comment on your definition of $\Bbb N$. There are many ways to define $\Bbb N$, and there's one quite close to your intuitive understanding of what $\Bbb N$ is. The axioms are called the Peano axioms, and I'll state them here:

  
*
  
*$1$ is a natural number ($1\in\Bbb N$)
  
  
  Then we define an $S(n)$ function (known as the successor function, but know that although intuitively, this should be $n+1$, the axioms say nothing about what $+$ means, and only uses $S$ to define the natural numbers)
  
  
*
  
*For all natural $n$, we know $S(n)$ must be a natual number ($n\in\Bbb N\implies S(n)\in\Bbb N$)
  
*For all natural $n$ and $m$, $n=m$ if and only if $S(n)=S(m)$ (that is to say, $S$ is an injection).
  
*For every natural $n$, $S(n)=1$ is false (that is, no natural number has $1$ as successor).
  

Now the important thing to note at first, is that the axioms say nothing about $2$, $3$, $1000$, or any other "numbers" (strings of digits); it merely shows us the structure of the natural numbers. To make it easier for ourselves, people've thought up symbols for each of these elements to be able to express them, but we could just as well give them other symbols. We could even give one symbol to each natural number, while maintaining the structure (for example, we can call $23$ the symbol $\circ$ and $24$ the symbol $\square$ and then $S(\circ)=\square$). As long as we follow the rules that are the axioms, we are still working with the same set, $\Bbb N$, but simply with a different representation of it. Now if we were to give each number its own symbol, it makes total sense that each number has a finite amount of "digits" - the word "digit" doesn't even mean anything here, since each number is one symbol anyway. See how we can have infinitely many natural numbers, while none has an infinitely long representation?
Furthermore, when you write $141258173412873\cdots$, you're not writing down an infinite number, since it must end somewhere (at the decimal dot), and since it starts, it cannot be infinite. But then you might ask, but what about $\cdots 141258173412873$? Is that a natural number? The question I would ask you then is: which one do you mean? If you were to write numbers this way, you cannot have a unique representation of each one. Also (going back to the axioms), how many times would you have to do $S$ on $1$ to get to that number? The axioms only state things about doing $S$ a finite amount of times, and your number would then require to do $S$ exactly $\cdots 141258173412872$ times, but this is not finite!
I think what you were stuck on before (and hopefully aren't anymore, after my and many other's attempts to explain it) is to wrap your head around the concept of infinity. How can you have infinitely many, but all finite numbers? It takes time and a lot of thought to get familiar with it, but I'm sure you have the motivation and the mind to do this. My main point is, don't think about natural numbers (or numbers from any set) as strings of digits, but rather as mathematical objects defined by some rules (axioms). The strings of digits are merely a common representation of them.
A: The set $\mathbb{N}$ of natural numbers is usually defined using the successor function: for natural number $n$, $S(n) = n+1$. We then define the natural numbers as follows:


*

*$0 \in \mathbb{N}$

*For all $n \in \mathbb{N}$, $S(n) \in \mathbb{N}$. 


So for your "number" $(141258173412873....)$ to be in $\mathbb{N}$, we would have to have a natural number $m$ so that $S(m) = (141258173412873....)$. Try to find this number and you see that it cannot exist, since "adding $1$" makes no sense when there is no $1$'s place to add the $1$ to.
A: You can prove by induction that $ N < 10^N$ is true for any natural number N. Then N has at most N digits, a finite number. Well, you have to believe that natural numbers are finite, but since they all are obtained by adding 1 to 1 a finite number of times, I think we agree at this point.
