Number of positive integers that do not contain the digit $9$ 
Prove that if $f(n)$ is the number of positive integers $k \leq n$ that do not contain the digit $9$, then $f(10^m) = 9^m$ for all $m \geq 1$.

I was thinking of using induction but then how do we determine the number of integers between $10^k$ and $10^{k+1}$ with no $9$ digit?
 A: This is the same as the number of non-negative integers $\le 10^m-1$ that have no $9$'s. 
Any integer from $0$ to $10^m-1$ can be thought of as a string of $m$ digits by "padding" numbers with fewer than $m$ digits with initial $0$'s. For example, if $m=4$, we can think of $12$ as $0012$.
So we want to count the digit strings of length $m$ that have no $9$'s.
The first digit can be chosen in $9$ ways. For every such choice the second digit can be chosen in $9$ ways. So the first two digits can be chosen in $9^2$ ways. For every such choice, the third digit can be chosen in $9$ ways. So the first three digits can be chosen in $9^3$ ways. And so on (here we have a hidden induction). So the $m$ digits can be chosen in $9^m$ ways.
A: 
I was thinking of using induction 

Good idea.  Go with that.

but then how do we determine the number of integers between $10^k$
   and $10^{k+1}$ with no 9 digit?

You likely mean those within $\{10^k+1, .., 10^{k+1}\}$ , since we don't wish to include the end of the last interval to avoid overcounting.
You are counting all of the $k$ digit strings where the first digit is selected from $\{1,2,3,4,5,6,7, 8\}$ and each of the other $k-1$ are selected from $\{0,1,2,3,4,5,6,7,8\}$; noting that the selection is independent ("allows repetition").  Then exclude that consisting of one and $k$ zeroes, but include the one consisting of one and $k+1$ zeroes.

 $$f(10^{k+1})-f(10^k) = 8\cdot 9^{k-1}$$

Now, you can continue with your induction proof.
