$$\begin{align*} & \because n(n-1)(n-2)(n-3) \cdots \lceil n/2 \rceil \leq (\lceil n/2 \rceil)^{\lceil n/2 \rceil} \\ & \therefore (\lceil n/2 \rceil)^{\lceil n/2 \rceil} \leq n! \\ & \because \log(\frac{n}{2})^{\frac{n}{2}} = \frac{n}{2}\log(\frac{n}{2}) \\ & =\frac{n}{2}(\log n -\log 2)=\frac{n}{2}\log n-\frac{n}{2}\log 2\\ & =\frac{n}{2}\log n-\frac{n}{4}\log 4 \geq \frac{n}{2}\log n-\frac{n}{4}\log n = \frac{n}{4}\log n\\ & \therefore \log(n!) = \Omega(n\log n) \\ & \because n! \leq n^{n} \\ & \therefore \log(n!)=\sum_{i=1}^{n}\log i \leq \sum_{i=1}^{n}\log n = n \log n \\ & \therefore \log(n!) = \mathcal{O}(n\log n) \\ & \therefore \log(n!) = \Theta(n\log n) \end{align*}$$