Factorial—Googol problem help So over the past week, I encountered this interesting problem in which I need help with, so any help is highly appreciated!
Problem: What it is least integer $n$ such that $n!$ is greater than googol?
My work so far: We are trying to find the minimal n where $n!\ge 10^{100}$. With this, we can log both sides and get $\log _{10} n! \ge 100$. Now we can split the LHS into $\log (n) + \log (n-1) +... + \log (1)$ to get $\log (n) + \log (n-1) +... + \log (1) \ge 100$.From here we dissect the LHS. We can take the nearest integer to each individual term. To know how many 2' s we'll have and etc., we'll have to find the maximum n in which $[\log(n)] = 1$ and the max n when $[\log(n)] = 0$. We find that n is 31, 3. Now we can set up the equation $2(n-32)+31-4+1+0 \ge 100$. We get n=68. However, this is not the exact answer, I know, however it is around there.
 A: You can use Stirling's approximation, which will still leave some calculation.  We have $$n!\approx \frac {n^n}{e^n}\sqrt{2 \pi n}\\
\log(n!)\approx n \log n -n +\frac 12\log(2 \pi n)$$
Maybe you know that $\log 10 \approx 2.30$, so we want $\log n! \gt \log 10^{100} \approx 230$.  I would start by ignoring the last term (we can make the correction later, when we are close), so we want
$$n (\log n-1) = 230\\n = \frac {230}{\log n-1}$$
which is nice for fixed point iteration if we have a calculator or choose $n$s that we can compute easily.  I know more base $10$ logs than natural logs, so I will write this as
$$n = \frac {230}{2.30 \log_{10} n-1}$$
If we make our first guess $100$, the log is $2$ and our next guess is $50.$  I know $\log_{10}5 \approx 0.70$ so our third guess is
$$n_3=\frac {230}{2.30 \log_{10} 50-1}\approx \frac{230}{2.30 \cdot 1.70-1}=\frac {230}{2^2-0.09-1}=\frac{2.30}{2.91}\approx 79$$
Then knowing $\log_{10}2 \approx 0.301, \log_{10}8 \approx 0.903$
$$n_4=\frac {230}{2.30 \log_{10} 79-1}\approx \frac {230}{2.30\cdot 1.90-1}\approx \frac{230}{3.37}\approx 68$$
and we are getting close.  If I truly had to not use a calculator, I would note that the $\frac 12\log{2 \pi n} \approx \frac 12\log(450)\approx 3$ and subtract that from the average of $68$ and $79$, getting about $70$ and call it good.  A final check with Alpha shows this is very close, but I would feel it was within $3$ either way based on the rough work.
