Showing log(n!) ≤ nlog(n) is easy using properties of log. For log(n!) = Ω(nlog(n)) use a bound related to Stirling's approximation: .
For b), simplify both sides using the fact that ln(x) and e^x are mutually inverse.
I assume you know calculus. The following are very often useful:
Let and be non-negative functions. Assume
1. If then and so also and .
2. If L=0, then , but (and so
Both 1) and 2) are easy consequences of the definition of limit and the orders.
Recall and so
By approximating Riemann sums and the fact that ln is increasing,
With a little more argument, you can see that
So by 1),