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For a positive integer $n$ define $S_n=1!+2!+\cdots +n!$. Prove that there exists an integer $n$ such that $S_n$ has a prime divisor greater than $10^{2012}$ .

Let's prove a more general result:

For a prime $P$ ,define $C_p = \{x : P^x \mid S_n , $ for some $n \in \mathbb N \}$, then $C_P$ is a bounded.

Proof: Assume , $C_P$ is not bounded for a prime $P$. Then -

Claim 1 : $P^x \mid S_{px}$ ,for any $x \in \mathbb N $
Proof: As $C_P$ is unbounded ,there exist integers $y$ & $Z$ such that $y\in C_P,y >x, Z>PX$ and $ P^y \mid S_Z $. Now if a integer $K> PX$ then $P^x \mid K!$. So $P^x \mid P^y \mid S_Z \Rightarrow P^x \mid S_{px}$

Claim 2: $P^{x+k} \mid S_{px}$ ,for any $x \in \mathbb N,k\in \mathbb N_0 $
Proof: Lets prove by induction on $K$.By claim 1 it is true when $K=0$ and $x \in \mathbb N$. Assume it is true for $K=L$ and any $x \in \mathbb N$ . So, $ P^{L+1} \mid S_{p}$ .Clearly $ P^{x+L} \mid { S_{P(x+1)} - S_{PX} }$.

So, $\frac{(PX+1)! +(PX +2)! + \cdots +(PX+P)!}{ P^{x+L}} \equiv \frac {(PX)!}{P^x}.(\frac {1!+2!+\cdots +P!}{P^L}) \equiv \frac {(PX)!}{P^x}. \frac {S_P}{P^L} \equiv 0 (mod $ $ P) $

So,$ P^{x+L+1} \mid { S_{P(x+1)} - S_{PX} }$. But $ P^{x+L+1} \mid S_{P(X+1)} \Rightarrow P^{x+L+1} \mid S_{PX}$ ,which completes the induction.

Claim 2 gives a contradiction which completes the main proof .
So, by this result Infinitely many primes divides $S_n$ .