How prove $a_{n}=1!+2!+\cdots+n!$ contains infinitely many prime factors show that $$(n+1)b_{n+1}-b_{n}=n+1,b_{1}=1,a_{n}=n!b_{n}$$,then the sequence $\{a_{n}\}$ contains infinitely many prime factors
my idea:since
$$(n+1)b_{n+1}-b_{n}=n+1$$
$$\Longrightarrow (n+1)!b_{n+1}-n!b_{n}=(n+1)!$$
so we easy find
$$a_{n}=n!b_{n}=1!+2!+\cdots+n!$$
then this problem equivalent to prove
$a_{n}=1!+2!+\cdots+n!$ contains infinitely many prime factors. But I can't prove it, thank you everyone.
 A: The key idea is to show that $a_n$ grows fast and if $a_n$ has only finitely many prime divisors, then the exponents have to grow very fast, which is impossible for this particular sequence. More precisely, let us denote by $d_p(n)$ the maximal power of $p$ that $n$ is divisible by. 
Now, take any prime divisor $p.$ If $d_p(a_n)<d_p((n+1)!),$ then $d_p(a_{n+1})=d_p(a_n)$ and in general $d_p(a_m)=d_p(a_n)$ for all $m\ge n.$ So the power of $p$ that $a_k$ is divisible by is bounded by the universal constant and we are not interested much in such primes as far as grows of the sequence is concerned.
Now, take any prime $p_i,$ $1\le i\le k$ such that $d_{p_i}(a_n),$ $n\ge 1$ is unbounded for any $1\le i\le k$. Then, for all $n\ge 1,$ $d_{p_i}(a_n)\ge d_{p_i}((n+1)!).$ Observe, that if $d_{p_i}(a_n)=d_{p_i}((n+1)!)$ for all $p_i$ then $1!+2!+3!+...+n!> n!$ and $a_n$ has to be much smaller than $(n+1)!$ because $(n+1)!$ has many more divisors than $p_1,p_2...p_k$ when $n\to\infty.$ This can easily be made more precise by claiming that $(n+1)!\ge 2^{n/2}{p_1^{d_{p_1}((n+1)!)}p_2^{d_{p_2}((n+1)!)}....p_k^{d_{p_k}((n+1)!)}}.$
So for some of $p_i$s, $d_{p_i}(a_n)$ has to be much bigger then $d_{p_i}((n+1)!).$ But $d_{p_i}(a_n)>d_{p_i}((n+1)!)$ only if $d_{p_i}(n!)=d_{p_i}(a_{n-1})$ and we can repeat the previous arguments with $a_{n-1}$ and $n!$ instead.
This completes the proof. 
A: Let $S_n=1!+2!+ \cdots +n!$.
Assume that the set $A$ of prime divisor of $S_n$ is finite. Let $p \in A$ then we find that for sufficient large $n \equiv -1 \pmod{p}$ then $p \mid S_n$. If $\nu_p \left( S_{n-1} \right) \ne \nu_p \left( n! \right)$ then we find that $\nu_p \left( S_i \right) = \min \left \{ \nu_p \left( S_{n-1} \right), \nu_p \left( n! \right) \right \}= \text{const}$ for all $i>n$. If this is true for all $p \in A$ then we find that $S_n$ is bounded, a contradiction.
Thus, this follows there exists primes $p_1,p_2, \cdots , p_k \in A$ such that $\nu_{p_i} \left( S_{n-1} \right) = \nu_{p_i}(n!)$ for $n \equiv -1 \pmod{p_i}$. Hence, $\displaystyle S_{n-1}=K \cdot \prod_{i=1}^k p_i^{\nu_{p_i}(n!)}=K \cdot \prod_{i=1}^k p_i^{\nu_{p_i}((n-1)!)}$ for all $n \equiv -1 \pmod{p_i} \; \forall i=\overline{1,k},$ where $K$ is a constant. Hence, we can pick a large enough $n$ such that $S_{n-1}<(n-1)!$, a contradiction.
Thus, set $A$ of prime divisor of $S_n$ is infinite. 
