Prove that if n is a natural number, then $\frac{\sigma(n!)}{n!}$ $\ge$ $\sum_{i=0}^{n}{\frac{1}{i}}$
I thought I would have to do something like showing $\frac{\sigma(n!)}{n!}$ $\ge$ $\sum_{i=0}^{n}{\frac{n!}{n! * i}}$ and then canceling for each term on the right side and left side to show the left side was greater, but I can't get any further.