How to show $\sum\limits_{i=1}^n\log\left(\frac{n}i\right)= \Theta(n)$?

You did it all by yourself, really:
Stirling's formula shows that $n^n/n!\sim(2\pi n)^{-1/2}\mathrm e^{n}$, hence $\sum\limits_{i=1}^n\log(n/i)=\log(n^n/n!)$ (your post) and $$\log\left(n^n/n!\right)=n-\frac12\log(2\pi n)+o(1)=n+o(n)=\Theta(n).$$

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Edit (without Stirling's formula)
Call $x_n=\sum\limits_{i=1}^n\log(n/i)=n\log(n)-\sum\limits_{i=1}^n\log(i)$, hence $$ x_{n+1}-x_n=(n+1)\log(n+1)-n\log(n)-\log(n+1)=n\log(1+1/n). $$ Since $\log(1+u)=u+o(u)$ when $u\to0$, $x_{n+1}-x_n=1+o(1)$ when $n\to\infty$. This implies that $x_n=n+o(n)=\Theta(n)$.