What's the first digit from left of $(2016)!$
I tried to use the Stirling formula $$n!\approx \dfrac{n^n}{e^n}\sqrt{2n\pi},$$ but I only could find the number of the digits, I didn't get the first digit. So how to calculate by hand that the first digit is $2?$
use WA:2325849581803\cdots
Apply $\log_{10}$ to $n!\approx \dfrac{n^n}{e^n}\sqrt{2n\pi}$ and get $$ \log_{10}(n!) \approx n (\log_{10}(n)-\log_{10}(e))+\frac12(\log_{10}(2 \pi)+\log_{10}(n)) $$ Evaluate this for $n=2016$ and get $$ \log_{10}(2016!) \approx 5788.36656 $$ The exact value is $5788.36658\cdots$ but two decimals are enough: $$ \log_{10}2 \approx 0.3 < 0.36 < 0.4 < \log_{10}3 $$ and so the first digit is $2$.
This computation can be done with a calculator or in floating point with a computer (which is what I've used), but Stirling's approximation can't be used directly because it overflows to infinity.
We can trust $\log_{10}(2016!) \approx 5788.36$ because the next term in the (alternating) Stirling series for $\ln n!$ is $\dfrac{1}{12n}$, which is $0.00004\cdots$ for $n=2016$. Multiplying by $\log_{10}(e)$, this gives a maximum error of $0.0002\cdots$ for $\log_{10}(2016!)$.