The relation between the number of $0$s which are at the end of $3^{n!}-1$ and that of $n!$ Let $a_n,b_n$ be the number of $0$s which are at the end of $3^{n!}-1,n!$ in the decimal system respectively. I found that $a_n=b_n+1$ holds for $n=4,5,\cdots, 10$. Then, my questions are...

Question 1 : Does $a_n=b_n+1$ hold for every $n\ge 4\in\mathbb N$?
Question 2 : If the answer for question 1 is no, then does $\lim_{n\to\infty}(a_n/b_n)$ exist?

We know that $b_n=\sum_{k=1}^{\infty}\lfloor n/5^k\rfloor$, but I don't have any good idea to treat $3^{n!}-1$. Can anyone help?
 A: Your conjecture is true. We have the following celebrated result in math olympiads:

Theorem. (Lifting The Exponent) If $p$ is an odd prime, $n\in\mathbb N$, $a,b\in\mathbb Z$ such that
  $p\mid a-b$ but $p\nmid a,b$, then
  $\nu_p(a^n-b^n)=\nu_p(n)+\nu_p(a-b)$, where $\nu_p(x)$ denotes the
  number of factors $p$ in the prime factorisation of the integer $x$.

Proof. See here, Theorem 1. (As you can see, it is a very elementary result.)
How does this apply here? Note that $5\mid 3^4-1$. As soon as $4\mid n!$ (that is, $n\geqslant4$) we have $\nu_5(3^{n!}-1)=\nu_5(3^4-1)+\nu_5(n!/4)=1+\nu_5(n!)$.
Now, what about the factors $2$? You may guess that $3^{n!}-1$ has a lot more factors $2$ than $5$. Indeed: Euler's theorem says: $2^k\mid3^{2^{k-1}}-1$ for $k\in\mathbb N$.
So, if $m=\nu_5(n!)$, then $m\leqslant\nu_2(n!)$ so $2^m\mid n!$, hence $3^{2^m}-1\mid3^{n!}-1$. Euler's theorem says $2^{m+1}\mid3^{n!}-1$, meaning that $\nu_2(3^{n!}-1)\geqslant m+1$: we've shown that $3^{n!}-1$ has at least as many factors $2$ as factors $5$.
Summarizing, since the number of trailing $0$'s of an integer $x$ is $\min(\nu_2(x),\nu_5(x))$, we have shown that $3^{n!}-1$ has one more zero than $n!$ has.
A: Just that you can check much higher; you know how to find $b_n.$ Then calculate 
$$  3^{n!} - 1 \pmod {2^{b_n + 2}},   $$
$$  3^{n!} - 1 \pmod {5^{b_n + 2}}.   $$
I don't see much reason that (1) should hold forever, but up to 100 would be impressive. 
There are topics sort of like this where there is agreement for an initial set but later disagreement. The superior highly composite numbers and the colossally abundant numbers agree on something like the first 15 terms, don't quite remember. And they stay roughly similar forever, they are approximately the least common multiple of the numbers from $1$ to some positive $N.$ But there are small disagreements, more pronounced as everything gets bigger.  
