Least value of $n$ such that $99^n$ begins with a $8$ 
MyApproach:
I have to calculate all powers of 99 and see when the number starts with 8.
Solving this step by step 
$99^1$=$99$
$99^2$=$9801$
Similarly for others I get n=11 

Is there Any other way I can calculate faster.Please help me if I am wrong?

 A: Outline, Not Rigorous, No Calculator
This is not rigorous, but this is what I will do if I were asked such a question in an exam.
We want $(100-1)^n = 100^n - \binom {n}{1} 100^{n-1} + \ldots$.
These are really the terms that contribute to the answer, so definitely $n \lt 10$ is not possible. Even for $n = 10$, we get a number starting with '9' here and the subsequent term is +ve and everything together is going to have a positive effect.
This leaves us with $\color{blue}{n = 11}$ which I have verified as the correct value.
More Intuition
We know that $99 \times 99 = 9801$. Now when we further multiply, the essential digits come from $98 \times 99$ which is $9702$ (ignore the 99 at the end). One more multiplication with $99$ and the product will start with $96 \ldots$. This goes on and at $\color{blue}{n = 11}$, we get the value $89\ldots$. (This was essentially explained using the binomial expansion above) 
A: Let's explain how this problem can be made into an easier problem with logarithms.
For $99^n$ to start with the digit $8$, we need $9\times10^x>99^n\geq8\times10^x$.
Taking logs on this inequality gives:
$x+\log_{10}(9)>n\log_{10}(99)\geq x+\log_{10}(8)$
Now we take the integer part:
$\log_{10}(9)>\{n\log_{10}(99)\}\geq\log_{10}(8)$
