Prove there exist n that satisfies $S_n = 1! + 2! + .... +n!$ have prime divisor greater than $10^{2020}$. 
Prove there exists $n$ such that $S_n = 1! + 2! + .... +n!$ has a prime divisor greater than $10^{2020}$.

I started this question like this:
For every $n$, $S_n$ have prime divisors only in range $(1, 10^{2020}]$. If $S_{n-1}$ is divisible by $n$ every greater $S_n$ will be divisible too, so once we have all possible prime numbers in  $S_n$ only thing that can change is their power. I also know that $S_n$ can't be perfect power. Can someone give me a hint what to do next?
(Full solution will work too :) )
Thanks in advance!  
 A: Suppose that no $S_n$ has a prime divisor greater than $N = 10^{2020}$.
Let $n \ge N$, then if a prime number $p \mid S_n$, we have that $p \mid S_{n+1}$, too. To see that note that $S_{n+1} = S_n + (n+1)!$ and by assumption $p < n+1$. Therefore if $A(n)$ is the set of prime divisors of $S_n$ we have the following increasing chain:
$$A(N) \subseteq A(N+1) \subseteq A(N+2) \subseteq \cdots$$
However, $A(n) \subseteq [2,N] \cap \mathbb N$, and as the latter is a finite set we have that the chain above will have to stabilize. Therefore there exists $M \ge N$, s.t. $A(n) = A(m)$ for all $m,n \ge M$. So from now on we'll label this set $A(M)$
We now consider the powers of the primes in the prime factorization of $S_n$ for large enough $n$. Let $p \in A(M)$ and denote by $\nu_p(n)$ the highest power of $p$ that divides $n$. Then we have the following lemma:
Lemma: If $\nu_p(S_{n_p}) < \nu_p((n_p+1)!)$, for some $n_p \ge M$, $\nu_p(S_{m}) = \nu_p(S_{n_p})$ for all $m \ge n_p$
The proof is trivial as $S_{n_p+1} = S_{n_p} + (n_p+1)!$ and we have that $p^{\nu_p(S_{n_p}) + 1}$ divides the second term, but not the first. Then just induct, as $\nu_p(S_{n_p+1}) = \nu_p(S_{n_p}) < \nu_p((n_p+1)!) < \nu_p((n_p+2)!)$. 
If for every $p \in A(M)$ we can find such $n_p$, then $S_n = S_{n+1}$ for large enough $n$, which is an obvious contradiction. So suppose $\exists p \in A(M)$, s.t. such $n_p$ doesn't exist. Consider $n = kp - 2$, s.t. $n \ge M$. If we have that $\nu_p(S_n) > \nu_p((n+1)!)$ for some $n \ge N$. Then as above we have that $\nu_p(S_{n+1}) = \nu_p((n+1)!) < \nu_p((n+2)!)$, where the strict inequality follows as we have at least one more $p$ factor from $n+2 = kp$. Therefore we must have $\nu_p(S_n) = \nu_p((n+1)!)$. But as $p \nmid n+1 = kp - 1$ we have that $\nu_p((n+1)!) = \nu_p(n!)$
Now using the Chinese Remainder Theorem we can find a natural number $n \ge M$ s.t. $n \equiv -2 \pmod p$ for all $p \in A(M)$ for which such $n_p$ doesn't exist. Then from the last paragraph we have that $\nu_p(S_n) = \nu_p((n+1)!) = \nu_p(n!)$ for all such $p$. In fact using the Lemma by taking $n$ to be large enough solution of the system of congruences we have $\nu_p(S_n) \le \nu_p(n!)$ for all $p \in A(M)$  As $A(M)$ contains all the prime divisors of $S_n$, this means that $S_n \le n!$, which is an obvious contradiction, as $S_n > n!$ from the definition.

As you might have guessed the number $10^{2020}$ is rather arbitrary and the above proof can be easily generalized to show that the set $\{p \mid p \text{ is a prime divisor of } S_n \text{ for some } n \in \mathbb N\}$ is unbounded.
