Number of digits in the number $N=(1.6 \times 10^{32})!$

Question

I am trying to find the number of digits in $$N=(1.6 \times 10^{32})!$$ where ! denotes Factorial. I have no idea how to proceed, please help me.

Answer 1

I can give an approximate answer. For large numbers (and $10^{32}$ certainly qualifies), Stirling's approximation holds: $$n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n.$$ Thus, the number of digits of large factorials is approximately $$\text{dig}(n!)\approx\log_{10}(n!)\approx\frac 12\log_{10}(2\pi n)+n\log_{10}(n/e).$$ In the case that $n=1.6\times 10^{32}$, I get $5.08317\times 10^{33}$.