How can I show that $\sum_{i=1}^n \frac{1+\log i}{n} = \Theta(\log n)$

Question

How can I show that $$\sum_{i=1}^n \frac{1+\log i}{n} = \Theta(\log n)$$

What I've tried: $$\sum_{i=1}^n \frac{1+\log i}{n} = \frac{1}{n^2} \sum_{i=1}^n {\log i} = \frac{\log(n!)}{n^2} = \frac{\log n^n}{n^2} = \frac{n\log n}{n^2} = \frac{\log n}{n}$$

Answer 1

Hint: $$ \sum_{i=1}^n \log i \leq n \log n $$