Divisors of n! including 1 which sum to n!

You can do this by induction. For $n=3$, you already observed that $1+2+3$. Now take $n>3$, then $(n-1)!$ is sum of some of its divisors, with $1$ included among them. By multiplying this equality by $n$, you'll get sum of divisors of $n!$, summing to $n!$, and the smallest of them is $n$. Now we can just put $1+(n-1)$ instead of $n$ and we're done.