I know the answer is $198$.
I realise that if $\log _{10}\left( x\right) =y$, the number $x$ has $\lfloor y\rfloor -1$ digits
So I tried $\log ^{\ }_{10}\left( 99^{99}\right) $ = $\log _{10}\left( 100\left( 1-\dfrac {1}{100}\right) \right) $ = $198 + 99\log_{10}\left( 1-\dfrac {1}{100}\right) $ , then I do not know how to proceed. I guess using this method amounts to finding a good approximation to $\log _{10}\left( 99\right)$
I would also be interested to know how one can solve this with the binomial theorem: $\left( 100-1\right) ^{99}$
Best Answer
In order to show that $99^{99}$ has $198$ digits, we need to show that
$$10^{197}\le99^{99}\lt10^{198}$$
The second inequality is obvious, since $99^{99}\lt100^{99}=10^{198}$. So it remains to prove the first inequality.
Toward this end, recall that
$$\left(1+{1\over n}\right)^n\lt e$$
for any $n\ge1$. We'll take for granted also the (generous!) inequality $e\lt10$. Thus
$$\left(100\over99\right)^{99}=\left(1+{1\over99}\right)^{99}\lt e\lt10$$
so $100^{99}\lt10\cdot99^{99}$. Since $100^{99}=10^{198}=10\cdot10^{197}$, the inequality $10^{197}\lt99^{99}$ now follows.