How to prove $10^n$ is the smallest $n+1$ digit number? It seems obvious that $10$ is the smallest 2 digit number and $100$ is the smallest 3 digit number.
In fact, it seems a little obvious to me that $10^n$ is the smallest $n+1$ digit number.
But how do I prove this without any assumptions?
 A: Well, one should start out by looking at what a decimal representation is: It is a representation of a number $x$ as a sum of powers of ten multiplied by digits. That is, a decimal representation with $n$ digits is a sum of the form
$$x=a_0+10\cdot a_1+100\cdot a_2+\ldots+ 10^{n-1}\cdot a_{n-1}$$
where each of the $a_i$ an integers between $0$ and $9$.
Now, it's pretty easy to see that the largest number with an $n$ digit representation is the one where all of $a_i$ are as big as possible - that is, they're all $9$. Thus the largest $x$ with an $n$ digit representation is
$$x=9+10\cdot 9 + 100\cdot 9 + \ldots + 10^{n-1}\cdot 9 = \underbrace{99\ldots 99}_{n\text{ times}}=10^n-1$$
One can prove by induction that all smaller positive numbers are indeed representable with $n$ digits. Obviously, we have that $10^n$ requires at least $n+1$ digits to represent and, given that all smaller numbers require only $n$ digits, it follows that $10^n$ is the smallest number requiring $n+1$ digits.
(Note that I use the terminology of requiring $n+1$ digits, since I can totally write $1=0001$ as a four digit representation of $1$. It's just not the smallest representation)
A: In order to prove something, you need to define it. However, 99% of mathematics is independent of the representation of numbers, so in order to prove this the easiest way is to invent a property of $x$ that corresponds to its number of digits. This is the output of the function $$f_b(x)=\lfloor\log_b(x)\rfloor+1$$
This function tells you how many digits the (positive real) number $x$ has in base $b$. It's a simple mater to now prove that $$f_b(b^n-\varepsilon)<n+1=f_b(b^n)$$ by showing that this function is strictly increasing.
A: By definition the number $N$ written as $a_n\;a_{n-1}\;...\;a_0$ in base 10 (where $a_k$ are its $n+1$ decimal digits) represents the value $a_n 10^n + a_{n-1} 10^{n-1} + ... a_1 10 + a_0$. Assuming the first digit $a_n \ne 0$ it follows that the number $N \ge 10^n$.
