Prove $2^{29}$ has exactly 9 distinct digits There is a problem that say: The number $2^{29}$ has exactly 9 distinct digits. Which digit is missing?
It is an Olympiad problem and I see it solved by using remainder modulo 9.
$2^{29}=536870912$  interesting!
My Question: Can we find any mathematically way to show $2^{29}$ has exactly 9 distinct digits?

That $2^{29}$ has exactly nine distinct digits is a coincidence, and a fairly appropriate one. One (a very good one) mathematical proof that $81619^2$ contains only two distinct digits, is calculating it : $81619^2 = 6661661161$. Even though sometimes these do yield patterns that are interesting, my personal take is that this is a one-off, a coincidence and a nice triviality to put in the back page of your notebook.
What we can do , is show that it has nine digits, and find the first few digits on both left and right. On the right, we would note that $2^{20} \equiv 1 \mod 100$ (checking remainders mod $25$), so the last two digits match with the first two digits of $512$, which give the last two digits as $12$. 
The first digit would come from the fact that $\log_{10}(2^{29}) = 29 \log_{10}2$. Noting that $\log_{10}(2) \approx 0.30103$ (for mathematical proof, use Taylor series) we get that $\log_{10}(2^{29}) \approx 10^{8.73}$, so from here you get that the number has $9$ digits, and the first digit is $10^{0.73}$, which again using a couple of Taylor series $e^{0.73 \ln 10}$ you can see to be $5$. Quite a bit of effort, but at least this method of deriving the digits is different from multiplying $2$ again and again.

Knowing that it has exactly $9$ distinct digits, we can actually calculate the missing digit using the fact that the sum of digits must be between $36$ and $45$. This can be located using the remainder modulo $9$.
Note that $2^3 = 8 \equiv -1 \pmod 9$. Therefore, $2^{27} \equiv (2^3)^{9} \equiv -1 \pmod 9$. Therefore, $2^{29} \equiv -4 \equiv 5 \pmod{9}$.
Therefore, the sum of digits of $2^{29}$ must leave remainder $5$ upon division by $9$, and be between $36$ and $45$. Thus, it is $41$. Now $45-41 = \color{blue}{4}$ is the missing digit.
