Is there a simple method to find or estimate how large $$\lfloor \lg{n!\rfloor}$$ is ? I'd like to find (or estimate) how much digits $2017!^{2017}$ has, or how much is big that number .
I tried for some little number and general form of $$n!^{n}=\overline{a_k...a_4a_3a_2a_1} \\ k=?$$ so I took logarithm of $n!^{n} \mapsto n\log(n!) $.
It is easy when $n$ is little number , but when go for a large number ...what we can do ? My question can be translate as $$n\sum_{i=1}^{n}\log(i)= ?$$
Stirling's approximation says that $$ n!\approx \sqrt{2\pi n}\left(\frac ne\right)^n $$ Taking logarithms on both sides should be easy enough.
$\lfloor \ln(n!)\rfloor=\sum_{k=2}^n\ln(k) + O(1)$. Since $\int_{k-1}^{k}\ln(t) dt\leq \ln(k)\leq \int_{k}^{k+1}\ln(t) dt$,
$$\int_1^n \ln(t)dt\leq \sum_{k=2}^n\ln(k) \leq \int_2^{n+1}\ln(t)dt$$
Since $\int \ln(x) dx = x\ln(x)-x$ and $\ln(n+1)\sim\ln n$, $$\sum_{k=2}^n\ln(k) = n\ln(n)+o(n\ln n)$$ hence $\lfloor \ln(n!)\rfloor=n\ln(n)+o(n\ln n)$.
However, this is only a partial answer as you need a more precise estimate for your problem.