How prove $a_{n}=1!+2!+\cdots+n!$ contains infinitely many prime factors

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.