How many digits are there in $99^{99}$?

Question

Question: How many digits are there in $99^{99}$?

My attempt:

Observe that $$99^{99} = (100\times 0.99)^{99} = 100^{99}\times 0.99^{99}.$$ Note that $100^{99} = 10^{198}$ has $199$ digits and $$0.99^{99} = \left(1-\frac{1}{100}\right)^{99} \approx e^{-1} \approx 0.37.$$ Therefore, there are $198$ digits in $99^{99}.$

Is my reasoning above correct?

Answer 1

Number of digits in $99^{99}$ are $1+[99\log_{10} 99]= [198.568]=198$ Here [.] denotes integer-part. So 198 digits is the correct answer.