Number of digits in a factorial sum $1!+2!+\cdots+100!$

Question

How do we find the number of digits in the sum $1!+2!+\cdots+100!$ , Options are given as:

1. $137$
2. $283$
3. $314$
4. $189$
5. none of these

So I tried to find individually number of digits but it wasn't effective.

Answer 1

$$\log_{10} (100!) <\color{blue}{ \log_{10} (1!+2!...+100!)} < \log_{10} (99(99!)+100!)< \log_{10}(2(100!))=\log_{10} (2)+\log_{10} (100!)$$

$$157.97...<\color{blue}{ \log_{10} (1!+2!...+100!)}<158.37..$$