Is there $n$ such that $n,n^2,n^3$ start with the same digit ($\neq 1)$ From the 2015 Moscow Mathematical Olympiad:

Is there some $n>2$ such that $n,n^2$ and $n^3$ start with the same digit (this digit being different from $1$)

Using a computer I found that $99$ fits the bill, but I'm wondering how this can be solved using only pen and paper.
I can't relate the decimal expansion of $n$ to that of $n^2$ and $n^3$ in the general case. Any ideas ? 
Also, is it possible to find all such $n$'s ? 

It is a bit unfortunate that I already saw that one answer would be 99, but it would probably one of my guesses since $99, 99^2, 99^3$ are very close to respectively $100, 100^2, 100^3$, so they would all start with the same digit, 9.
For example, if one had to find an $n$ where the first two digits of $n, n^2, n^3$ are the same, 999 would work, since they are all three even closer to $1000, 1000^2, 1000^3$.

Here is a list of numbers I found just by using this logic: 97, 98, 99, 966, 967, 968, 969, ... 997, 998, 999, 9655, 9656, ... , 9998, 9999, ...
In general, any number $k$ with $n$ digits such that $k > 10^n \cdot \sqrt[3]{0.9}$ works.
Note that $\sqrt[3]{0.9} \approx 0.96549$. This also means that there are infinitely many, and that they take about 3.5% of the positive integers. 
I suspect that this are all such numbers. 
