Let $T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}$. Given that $9^{4000}_{}$ has 3817 [digits](https://artofproblemsolving.com/wiki/index.php/Digit) and that its first (leftmost) digit is 9, how many [elements](https://artofproblemsolving.com/wiki/index.php/Element) of $T_{}^{}$ have 9 as their leftmost digit?
Lemma: For all positive integers n, there's exactly one n-digit power of 9 that does not have a left-most digit 9
(Not-so-rigorous) Proof: One can prove by contradiction that there must be at least either one or two n-digit power of 9 for all n.
If there is exactly 1 n-digit power of 9, then such a number $m$ cannot begin with 9 since that would result in $\frac{m}{9}$ also being an n-digits, hence a contradiction. Therefore, this single n-digit power of 9 must not begin with 9. In the case that there are 2 n-digit powers of 9, we have already discovered that one of them must begin with 9, let it be $9^k = m$. The integer $9^{k - 1} = \frac{m}{9}$ must then be an n-digit number that begins with $1$. 

Using this lemma, out of the 4001 powers of 9 in T, exactly 3816 must not begin with 9 (one for each digit). Therefore, the answer is $4000 - 3816 = \boxed{184}$ numbers have 9 as their leftmost digits.
~Edited by Mathandski (with help from a math god)